Basics of Calculus
In this calculus essentials series, I introduce and explain in detail the basic concepts that you will need to learn calculus.
Rates of Change
In the examples that are to follow I try to provide any explanations that seem necessary. I want to be clear and comment what I do and why I do it. I am hoping this method will help others understand. If it seems a little long or self-explanatory then I apologize. My goal is to be consistent and thorough. We are dealing with functions in this first chapter and the rate at which they change.
To get the most out of these examples I would encourage you to try the problem first by yourself. If you have issues then refer to how it is worked. Hopefully that, along with my comments, will fill in any gaps that you might have in your knowledge.
Problem 1
Find the average rate of change of the function over the given intervals. \[f(x)= 3x+3\] Using the intervals: [2,4] [-2,2]
Use the given function and the first interval.
You are evaluating the function at both points in order to find the difference.
Once you do that you are dividing by the difference of the interval length.
For the interval [2,4]
\[ \Longrightarrow \frac{f(4)-f(2)}{4-2} \]
Substitute in all of your values to find the rate of change.
\[ \Longrightarrow \frac{((3)(4)^3 +3)-((3)(2)^3 +3)}{4-2} = 84 \]
For the interval [-2,2]
\[ \Longrightarrow \frac{f(2)-f(-2)}{2+2}\]
Evaluate the equation with those values.
\[ \Longrightarrow \frac{((3)(2)^3+3)-((3)(-2)^3+3)}{2+2} = 12\]
Problem 2
Find the average rate of change of the $R\theta$ function over the given interval: [0,8]
\[ \Longrightarrow r \theta = \sqrt{3\theta+1} \]
Using this formula, we want to use the intervals given to us and evaluate them into the function. How this is done is simple. In the numerator we are using the intervals within the function itself. In the denominator, we just have the interval difference. Then we are just dividing the two.
\[ \Longrightarrow \frac{r \theta_2 - r \theta_1}{\theta_2 - \theta_1} \]
I will now go ahead and set up the equation.
\[ \Longrightarrow \frac{r (8) - r (0)}{8-0} \]
Now just substitute the intervals into the formula and start evaluating.
\[ \Longrightarrow \frac{\sqrt{3*8+1}-\sqrt{3*0+1}}{8-0} \]
This gives us a simple equation to solve.
\[ \Longrightarrow \frac {5-1}{8-0} \]
We are left with :
\[ \Longrightarrow \frac {1}{2} \]
This is your rate of change.
Problem 3
Find the slope of the curve at the given point and an equation of the tangent line at that point. $$(-2,-9)$$
\[ y=7-4x^2 \]
Approach this problem as change in y divided by change in x.
\[ \Longrightarrow \frac{(7-4(-2+h)^2) -( 7-4(-2)^2)}{h} \]
Distribute everything out.
\[ \Longrightarrow \frac{16h-4h^2}{h} \]
Now simplify and set h=0.
The slope at (-2,-9) is: 16
Now to find the equation of the tangent line. You use this formula for that:
\[ \Longrightarrow y-y_1 = m(x-x_1) \]
Substitute in your points and you get:
\[ \Longrightarrow y-(-9) = 16(x-(-2) \]
Simplify a bit.
\[ \Longrightarrow y+9 = 16(x+2) \]
Keep simplifying. This is the equation of the tangent line.
\[ \Longrightarrow y = 16x + 23 \]
Problem 4
Find the slope of the curve at the given point by finding the limiting value of the slope of the secants through the point. Also find an equation of the tangent line to the curve at the same point.
\[ y = x^3 - 6x \]
Point = (1,-5)
This is just another form of what we have done before. You have an equation and a point to go by. Put the point into the function and evaluate it.
\[ \Longrightarrow \frac{(1+h)^3 - 6(1+h) - (1)^3 6(1)}{h} \]
Now distribute. Pay careful attention to where everything goes as it is a lot to deal with.
\[ \Longrightarrow \frac{(1+3h+3h^2 +h^3-6-6h) - (-5)}{h} \]
Reduce it down some.
\[ \Longrightarrow \frac{h^3 + 3h^2 -3h}{h} \]
Simplify it now.
\[ \Longrightarrow h^2 + 3h - 3 \]
Then just assign h=0 and see what you have left. That is your slope of course as h approaches 0.
\[ \Longrightarrow = -3 \]
Now to find the equation of the tangent line.
We do the same process as before. We have the point-slope equation.
\[ \Longrightarrow y-y_1 = m(x-x_1) \]
Use your slope that you got above. Then put in your point when it asks for an x or a y.
\[ \Longrightarrow y-(-5) = -3(x-1) \]
Simply the equation.
\[ \Longrightarrow y + 5 = -3x + 3 \]
Move everything to one side of the equal sign. This is your equation of the tangent line.
\[ \Longrightarrow y = -3x - 2 \]
Limits
Limits of a function have been around in concept for a very long time. Mathematicians first started using the Limit concept somewhere around 400 years ago but our modern form of a Limit has only been used since the early 1800's or so. They are the foundation to learning derivatives and are subsequently used in modern Calculus quite often.
What Is A Limit
Lets say you are starting at 0,0 on this graph. That means that your X value =0 and your Y value =0. If you then move to some value to your right like X=4 then your X value =4. What does your Y value equal? Well that depends on whether you traveled in a straight line or some type of diagonal path up. Why does that matter?
Well, the Limit = the Y value!
So if you drew a crazy line that just went to X=4 but your Y value =7 then your Limit at X=4 =7. Does that make more sense now?
Wherever you end up on the X plane you will have some Y value as well. If you are still on Y =0 then your Limit =0.
So now that you should hopefully understand what the Limit is lets look at some problems.
Problem 1
\[ \tag{1}\lim_{x \to 3} f(x) \]
When you look at this graph think about what you see. You see the X-axis marked off and the Y-axis is too. Now what is the question asking? It is asking what is the limit of f(x) as X approaches 3. What is the limit then? The limit is the value that the Y-axis is when X is at or very near 3. When you look at the graph again you will see that Y = 0 so the limit is 0.
\[ \tag{2}\bbox[red,2pt]{\lim_{x \to 3} f(x) = 0}\]
Problem 2
What is
\( \tag{1}\lim_{x \to 9} f(x) \) :
This is the same graph as above and looking at it we can see our X and Y values clearly marked. Our question is asking what is the limit of f(x) as X approaches 9. Now we look at X=9 on the graph and then see what Y values are there. Do you see the issue yet? Yes there are two values for Y at X=9. These are one sided limits which we will get into another time but the thing to take away here is those two Y values are not equal. This means the limit in our problem does not exist.
\[ \tag{2}\bbox[red,2pt]{\lim_{x \to 9} f(x) = }\] Does Not Exist!
Problem 3
What is the
\( \tag{1}\lim_{x \to -2} ( -x^2 + 9x -1 ) \)
This is a limit problem that does not have a graph to look at but instead uses an equation. This equation is a simple polynomial. What can you do here? Well in math it is a good idea to try the simplest idea first to rule it out unless you have some reason not to. Lets try substitution here since that is barely algebra and it looks like it could work and give us a value. What we will want to do is substitute X=-2 into our equation. Why X=-2? That is because it is the point at which X is approaching.
\( \tag{2} ( -(-2)^2 + 9(-2) -1 ) \)
You should notice now that everywhere there was an (X) value in the equation I put in a (-2).
\(\tag{3}(-4 -18 -1)\)
Multiply the values together and remember to keep the signs right.
\(\tag{3}\bbox[red,2pt] {-23}\)
Problem 4
Find \(\tag{1}\lim_{y \to -11}(16-y)^4/3\)
Here we have another substitution problem where our variable is Y. It is a little different because our equation has an exponent. So lets take it step by step.
\(\tag{2}\lim_{y \to -11}(16 - (-11))^4/3 \)
Now we have substituted -11 for our Y variable. Then we simplify this a bit.
\(\tag{3}\lim_{y \to -11}(16 + 11)^4/3 \)
\(\tag{4}\lim_{y \to -11}(27)^4/3 \)
When dealing with an unusual exponent like this you can remember this little trick. Take your base number to the 4th power and then take the 3rd root of it. The 3rd root is also called the cube root. This is all the same as taking 27 to the \(4/3\) power.
Our little problem now becomes much easier and the solution is \(\tag{5}\bbox[red,2pt]{\lim_{y \to -11}(16 - y)^4/3 = 81}\)
Problem 5
Find \(\tag{1}\lim_{h \to 0}(\frac{6}{\sqrt{6h + 4} +4})\)
This is a nice problem because it combines a few different styles and techniques to finish. This one is still quick to finish but it will be the start of harder problems that look like this one.
We have a limit, a fraction, and and a square root to deal with so this will be fun while it lasts.
Now for this and any other problem involving a fraction, we can substitute as long as the bottom part of the fraction does not equal 0. We should be safe here by the looks of it so lets try that.
\(\tag{2}\lim_{h \to 0}(\frac{6}{\sqrt{6(0) + 4} + 4}) \)
Simplifying this now makes it easier to see what is happening.
\(\tag{3}\lim_{h \to 0}(\frac{6}{\sqrt{4} + 4}) \)
We still are not doing anything with the top but we are almost there. Lets finish with the bottom now.
\(\tag{4}\lim_{h \to 0}(\frac{6}{2 + 4}) \)
\(\tag{5}\lim_{h \to 0}(\frac{6}{6}) \)
Well this is shaping up nicely.
\(\tag{6}\bbox[red,2pt]{\lim_{h \to 0} = 1} \)
Problem 6
Find \(\tag{1}\lim_{h \to 0}(\frac{\sqrt{19h + 1 } -1 }{h}) \)
Ok different looking problem here. You can see we have the [h] on the bottom now. Since this is a fraction and we already know you can never divide by 0, we now know you must do something entirely different to solve this problem. Your main clue is that there is a square root on the top. Remember that square roots multiplied by themselves gives the value inside the root and the root disappears. However, what you do to the top will also have to be done on the bottom. This is called the conjugate method and it is used to solves problems like this.
\(\tag{2}\lim_{h \to 0}(\frac{\sqrt{19h + 1} -1 }{h}) * (\frac{\sqrt{19h + 1} + 1}{\sqrt{19h + 1} +1}) \)
Now we are just multiplying the left side by the right side. Watch your signs carefully here. Always good to keep everything spaced apart a lot so that you can see what your doing and avoid making mistakes.
\(\tag{3}\lim_{h \to 0} (\frac{19h + 1 -1}{{h}{\sqrt{19h+1}+1}})\)
Now that we have multiplied both sides together we can subsequently start simplifying the top and bottom of our equation.
\(\tag{4}\lim_{h \to 0}(\frac{19h}{{h}{\sqrt{19h+1}+1}})\)
Since we have an [h] on top and bottom being multiplied by another expression we can cancel those [h] out.
\(\tag{5}\lim_{h \to 0} (\frac{19}{\sqrt{19h+1}+1})\)
Substitute 0 for h on the bottom part of the equation.
\(\tag{5}\lim_{h \to 0}(\frac{19}{0+1+1})\)
We are basically done so lets just make it look nice and final. The answer is:
\(\tag{6}\bbox[red,2pt]{\lim_{h \to 0}=(\frac{19}{2})}\)\
Problem 7
Find \(\tag{1}\lim_{x \to 3}(\frac{x-3}{x^2-9}) \)
Here is another type of problem involving a limit however it just uses algebra to solve. You should be able to recognize instantly that the top and bottom parts of the equation are related to each. So break the problem up into parts and simplify from there.
\(\tag{2}\lim_{x \to 3}(\frac{x-3}{(x+3)(x-3)})\)
Since you will have an x-3 on both the top and the bottom you can just cancel them out in this situation. That will leave you:
\(\tag{3}\lim_{x \to 3}(\frac{1}{x+3})\)
Now you just apply the limit and you will have your answer.
\(\tag{4}\lim_{x \to 3}(\frac{1}{3+3})\)
This will give us a nice fraction as the answer.
\(\tag{5}\bbox[red,2pt]{\lim_{x \to 3}=(\frac{1}{6})}\)
Problem 8
Find \(\tag{1}\lim_{x \to 4} (\frac{x^2-2x-8}{x-4})\)
Here is another problem that is a fraction and it looks different than the one before you will notice. Actually, they should all look slightly different than the one before because there can be many variations. This is yet another variation with a polynomial. These problems are arranged this way to teach you the steps in troubleshooting a calculus problem and to help you recognize common forms of problems and what to do with them. If you are new to calculus just keep practicing until you instantly see what to do. Now here we go!
If you look at this problem it might seem weird but it is just a factoring calculus problem. You will obviously need to factor the top and hopefully that is clear to you. However, what beginners might not recognize is that the problem is already half factored for you. At this level of calculus problems are not going to be given to you that do not factor easily.
So, with that in mind, remember how the previous problems were dealt with? Yeah we factored and then cancelled out a top and bottom factor. We will do the same here. The bottom part of the fraction is one of the factors of the top part of the fraction.
\(\tag{2}\lim_{x \to 4}(\frac{(x+2) (x-4)}{x-4})\)
So see my point there how that works? We now have x-4 on both the top and bottom so we can just cancel them out.
\(\tag{3}\lim_{x \to 4}(x+2)\)
Now we just apply the limit.
\(\tag{4}\lim_{x \to 4}(4+2)\)
We now have a solution.
\(\tag{5}\bbox[red,2pt]{\lim_{x \to 4}=(6)}\)
Problem 9
Find the limit of \(\tag{1}\lim_{u \to 1}(\frac{u^3-1}{u^4-1})\)
There is a lot of factoring to do in this problem and you should be familiar with how to deal with numbers to the 3rd and 4th power. It is not hard at all though and furthermore it is good practice for harder problems later on. First, factor top and bottom and then see what we can do after that.
\(\tag{2}\lim_{u \to 1}(\frac{(u^2+ 1u +1)(u-1)}{(u^2+1)(u+1)(u-1)})\)
There we have both the top and the bottom parts of the fraction factored. Lets see what we can do with it now. We can cancel expressions on both top and bottom subsequently really make this simpler.
\(\tag{3}\lim_{u \to 1}(\frac{u^2 + 1u + 1}{(u^2 + 1)(u + 1) })\)
We are almost there now and we just need to substitute to apply the limit.
\(\tag{4}\lim_{u \to 1}(\frac{1^2 + 1(1) +1 }{(1^2 +1)(1 + 1)})\)
Now just do the simple math and we will have our answer.
\(\tag{5}\bbox[red,2pt]{\lim_{u \to 1} = 3/4 }\)
Problem 10
Find \(\tag{1}\lim_{x \to -6} (\frac{5-\sqrt {x^2 - 11}}{x+6})\)
Here is another nice problem but with a twist because now the square root is on the top. From looking at the top we can treat this as another conjugate problem. So lets multiple the top and bottom by the conjugate of the top.
\(\tag{2}\lim_{x \to -6} (\frac{5-\sqrt{x^2 - 11}}{x + 6}) * (\frac{5 + \sqrt{x^2 - 11}} {5 + \sqrt{x^2 - 11}}) \)
That is an ugly equation and set up so let us try to make it look a little nicer. Just multiple top and bottom and remember to keep the sings straight.
\(\tag{3}\lim_{x \to -6} (\frac{25 - (x^2 - 11)}{(x + 6)(5 + \sqrt{x^2 - 11})})\)
While this still looks messy and difficult to work with it is getting better. Keep simplifying things out step by step so you do not miss any signs or do something silly.
\(\tag{4}\lim_{x \to -6} (\frac{ -x^2 + 36}{(x + 6)(5 + \sqrt{x^2 - 11})})\)
Rearrange the top so that it makes more sense as to what your supposed to.
\(\tag{5}\lim_{x \to -6} (\frac{36 - x^2}{(x + 6)(5 + \sqrt{x^2 - 11})})\)
Now factor what you can and this problem start to make sense.
\(\tag{6}\lim_{x \to -6} (\frac{(6 + x)(6 - x)}{(x + 6)(5 + \sqrt{x^2 - 11})})\)
Now cancel the similar terms and lets see what is left.
\(\tag{7}\lim_{x \to -6}(\frac{6-x}{5 + \sqrt{x^2 - 11}})\)
Substitute and apply the limit now. We get a good integer answer.
\(\tag{8}\bbox[red,2pt]{\lim_{x \to -6} = (\frac{6}{5})}\)
Problem 11
Find the limit of \(\tag{1}\lim_{x \to 0}(\csc(x))\)
Now things are getting interesting because we now have limits of trig functions. Hopefully you remember your pre-calculus! These are simple though as you just rewrite the harder forms into something that you do recognize.
\(\tag{2}\lim_{x \to 0}(\frac{1}{\sin(x)} )\)
We know that \((\sin (0) = 0) \) but if you graph it or do a table of values you will see that values go wildly out of control as they get close to 0. That is our clue there.
\(\tag{3}\bbox[red,2pt]{\lim_{x \to 0} (\csc(x) = \infty)} \)
Problem 12
Find the limit of \(\tag{1}\lim_{h \to 0} (\frac{f(x + h) + f(x)}{h} )\) when \(f(x) = x^2\) and \(x = 7\).
\(\tag{2}\lim_{h \to 0} (\frac {(x + h)^2 -x^2}{h})\)
We just substituted in our functions and the value of x.
\(\tag{3}\lim_{h \to 0} (\frac{(7 + h)^2 - 7^2}{h})\)
We just plugged things in but it does take some attention to detail to keep things straight. We now have:
\(\tag{4}\lim_{h \to 0} (\frac{49 + 14h - 49}{h} ) \)
Just factor everything and watch expressions disappear again.
\(\tag{5}\lim_{h \to 0}(\frac{h(14 + h)}{h})\)
Factor out the \(h\) now so that we no longer have a fraction to deal with.
\(\tag{6}\lim_{h \to 0}(14 + h)\)
Now you just evaluate the expression and you get the answer.
\(\tag{7}\lim_{h \to 0}(14 + 0)\)
So our answer is :
\(\tag{8}\bbox[red,2pt]{\lim_{h \to 0} = (14)} \)
Definition of a Limit
The definition of a limit continues on in our learning limits and how they are applied. Specifically we are going to look at some proofs in this section. With a precise definition we can now prove many limit properties.
The \(\delta\) value will always depend on the \(\epsilon\) value. That is important to remember in order to understand what is happening in these limits. Once you find a value that works in your problem then smaller values of that \(\delta\) also work.
Problem 1
For the given function f(x) and values L,c, and \(\epsilon\) > 0 find the largest open interval about c on which the inequality |f(x)-L| < \(\epsilon\) holds. Then determine the largest value for \(\delta\) >0 such that 0<|x-c|<\(\delta\)\(\rightarrow\) |f(x)-L|<\(\epsilon\).
f(x) = \(\frac{1}{x}\), L= 0.125, c = 8, \(\epsilon\) = 0.015
Solve |f(x) - L| < \(\epsilon\) to find the largest interval containing c on which the inequality holds.
Write the inequality without absolute value. Substitute \(\frac{1}{x}\) for f(x), 0.1215 for L, and 0.015 for \(\epsilon\).
|f(x)-L| < \(\epsilon\)
|\(\frac{1}{x}\) - 0.125| < 0.015
-0.015 < \(\frac{1}{x}\) - 0.125 < 0.015
Simplify by adding 0.125 to all expressions.
-0.015 < \(\frac{1}{x}\) - 0.125 < 0.015
0.11 < \(\frac{1}{x}\) < 0.14
Simplify further by taking the reciprocals of all terms. Be sure to reverse the inequalities. Make sure you check that the rounded endpoints still allow the inequality |f(x)-L| < \(\epsilon\) to hold true.
9.0909 > x > 7.1429
Since the interval 7.1429 < x < 9.0909 is not centered on c=8, \(\delta\) is the distance from 8 to closer endpoint of the interval. The value of \(\delta\) is then 0.8571.
Problem 2
For the given function \(f(x)\), the point \(c\), and a positive number \(\epsilon\), find \(L=lim f(x)\). Then find a number \(\delta\) > 0 such that for all x, 0<|x-c|<\(\delta\)\(\rightarrow\)|\(f(x)-L\)| < \(\epsilon\).
\(f(x)=8-3x, c=4, \epsilon=0.03\)
Notice that \(f(x)\) is a linear function. Since a linear function is a simple polynomial function and it is defined for all \(x\), the limit \(L\) = lim is the value of \(f(x)\) at \(c\).
Evaluate the function at \(c=4\).
\(L=8-3(4)=-4\)
To find \(\delta\), begin by solving the inequality \(|f(x)-L|<\epsilon\) to find an open interval \((a,b)\) containing \(c\) on which the inequality holds for all \(x\neq c\).
Remove the absolute value sign and rewrite the inequality as a compound inequality.
\(|(8-3x)-(-4)| < 0.03\)
\(-0.03 < (8-3x) - (-4) < 0.03\)
Simplify the center expression and isolate the x-term.
\(-0.03 < -3x+12 < 0.03\)
\(-12.03 < -3x < -11.97\)
Then isolate \(x\) in the center. Notice that the direction of the inequalities has been changed.
\(4.01 > x > 3.99\)
Therefore, for \(x\) in the interval \((3.99,4.01)\), the inequality \((8-3x)-(-4)|<0.03\) holds. Now choose a positive value for \(\delta\) that places the open interval \((c-\delta,c+\delta)\) centered on \(c\) inside the interval \((3.99,4.01)\).
The distance to the first endpoint is \(4-3.99=0.01\).
The distance to the second endpoint is \(4.01 -4=0.01\).
The largest possible value for \(\delta\) is \( 0.01\).
Therefore, for all \(x\) satisfying \(0<|x-4|<0.01\), the inequality \(|(8-3x)-(-4)|<0.03\) holds.
Problem 3
Give an \(\epsilon - \delta\) proof of the limit fact.
\(lim_{x \to 0}(5x+1)=1\)
We begin by giving the precise meaning of a limit.
To say that \(lim f(x)=L\) means that for each \(\epsilon>0\) there is a corresponding \(\delta>0\) such that \(|f(x)-L|<\epsilon\), provided that \(0<|x-c|<\delta\).
To establish the proof, we first perform a preliminary analysis to find the appropriate choice of \(\delta\).
Let \(\epsilon\) be any positive number.
We must produce a \(\delta > 0\) such that \( 0<|x-0|<\delta \rightarrow |(5x+1) -1 < \epsilon\).
By simplifying the inequality on the right, we will determine the value of \(\delta\) needed for the inequality on the left.
\(|(5x+1)-1| < \epsilon \Leftrightarrow |5x| < \epsilon \)
\(\Leftrightarrow |5| |x| < \epsilon\)
\(\Leftrightarrow 5|x| < \epsilon\)
\(\Leftrightarrow |x| < \frac{\epsilon}{5}\).
Now we see that we should choose \(\delta=\frac{\epsilon}{5}\). We can now construct the formal proof of the statement \(\lim_{x \to 0}(5x+1)=1\).
Let \(\epsilon > 0\) be given. Choose \(\delta=\frac{\epsilon}{5}\). Then \(0<|x-0|<\delta\) implies the following chain of equalities and an inequality.
\(|(5x+1)-1| = |5x|\) Simplify inside the absolute value bars.
\( = |5| |x|\) Rewrite as product of absolute values.
\( = 5 |x| \) Simplify.
\( =5 |x-0|\) Rewrite expression inside absolute value bars.
\( < 5\delta\) Apply the condition \(0<|x-0|<\delta\).
\( = \epsilon \) Substitute \(\delta=\frac{\epsilon}{5}\).
The result of the above chain of equalities and an inequality is that \(|(5x+1)-1|< \epsilon\).
Therefore, we have proven that \(\lim_{x \to 0}(5x+1)=1\).
Conclusion
The definition of a limit has some nice proofs that can establish. If you go over these problems slowly the ideas behind limits will begin to emerge. While this section is short it is essential to understanding limits and beginning calculus in general.
One-Sided Limits
Our current idea of the limit definition goes back around 200 years ago by Bernard Bolzano. He was instrumental in several mathematical ideas. One of which was the definition of one-sided limits which I will be talking about today. Now on to some problems!
Problem 1
Use the graph to see if the following statements are true or false. Look at the graph carefully because it can be tricky to see what is really going on sometimes.
We can see there is a function \(y=f(x)\). There is also a small portion of the positive x-axis defined from [3 to 6].
A. True or False: \(lim_{x \to -3+}(f(x)=9\)
F(x) is your [y] value. X is of course your [x] value. Does [y] or [f(x)] get closer to 9 as [x] gets closer to 3 when coming from larger numbers than +3 ?
We can see that it sure does. The answer is [true].
B. True or False: \(lim_{x \to 0-}(f(x)=3\)
F(x) is your [y] value. X is your [x] value. The question is asking if [y=3] as [x] approaches 0 from the negative side of the [x] axis ?
It appears that [y=0] when [x=0] so the answer is false for this question.
C. True or False: \(lim_{x \to 0-}(f(x)\) = \(lim_{x \to 0+}(f(x)\) ?
This question is asking if the left limit is the same as the right limit. What does [y] equal when [x] is coming from the negative side? It equals 0. Now what does [y] equal as [x] is coming from the positive side? It equals 0 also. So both one-sides limits equal each other and the answer to the question is [true].
D. True or False: \(lim_{x \to 0}(f(x)\) exists?
We have basically answered this question already. When the limit is listed without directional signs(- or +) it is asking if the limit exists and is equal from both sides of the graph. It is equivalent to asking if the limit from the left side equal the limit from the right side? As mentioned before we have answered this. We did see that the limit from the left side equaled the limit from the right side. So the answer to this question is [true].
E. True or False: \(lim_{x \to 0}(f(x)=0\)
This is another version of the question above. Since there are no directional signs(- or +) it is implying the full limit. The full limit is each one-sided limit being equal to each other. The left hand limit=0. The right hand limit=0. Therefore the limits equal each other and both equal 0. So the answer is also true for this question.
F. True or False: \(lim_{x \to 3}(f(x)=9\)
F(x) is your [y] value. X is your [x] value. The question is asking if [y=9] whether [x] approaches 3 from the negative or positive side?
When [x] is coming from the positive side of the [x] axis things are pretty clear and defined. However, there is a hole in the graph when [x] approaches 3 from the negative side. Therefore our limits are not equal and the answer is [false].
G. True or False: \(lim_{x \to 6-}(f(x)=6\)
F(x) is your [y] value. X is your [x] value. This is asking if [y=6] when [x] approaches 6 from the negative side? We can plainly see that it does not. When [y=6] it looks like [x] is somewhere between 2 and 3. So the answer to this question is [false].
Problem 2
F(x) = 6-x when x < 2 and (x/2) +1 when x > 2.
Find \(lim_{x \to 2+}(f(x)\) and \(lim_{x \to 2-}(f(x)\).
This is what I call a conditional function. Certain sections of the function give different values depending on the condition.
We will start by looking at the one-sided limits. First lets look at the limit as [x] approaches 2 from the positive side of the x-axis. The expresison (x/2) + 1 is used to evaluate when x > 2.
The \(lim_{x \to 2+}(f(x)=((x/2)+1)\) = (2/2) + 1 = 2.
The expression 6-x is used to evaluate the expression when x < 2.
The \(lim_{x \to 2-}(f(x)=6-x\) = 6-2 = 4.
Does the limit exist? For a full limit to exist its one-sided limits have to exist and be equal. The two limits above that we evaluated are not equal. Therefore the limit does not exist.
Problem 3
Use the relation \(lim_{\theta \to 0}{\frac{\sin\theta}{\theta}}=1\) to find the limit of this function.
\(\frac{\sin \sqrt{2} \theta}{\sqrt{2} \theta}\)
As \(\theta\) gets closer to 0 the square root of \(\theta\) also gets closer to 0.
We now have : \(lim_{\sqrt{2}\theta \to 0}{\frac{\sin \sqrt{2} \theta}{\sqrt{2} \theta}}\)
So \(x = \sqrt{2} \theta \). Now substitute in the values.
\(lim_{x \to 0} \frac{\sin x}{x} =1 \). That is our answer using the relation.
Problem 4
Find the limit of this expression. \(lim_{y \to 0}{\frac{\sin 3y}{7y}}\)
Again we will use the theorem \(lim_{\theta \to 0} \frac{\sin \theta}{\theta} = 1 \)
To get the form that we want so we can solve easily we need to multiply the numerator and denominator by \(\frac{3}{7}\).
Once you see how everything cancels out you will see that your answer is also \(\frac{3}{7}\).
Problem 5
Use the relation \(lim_{\theta \to 0} \frac{\sin \theta}{\theta} =1 \) for the following one-sided limits problem.
\( f(x) = \frac{2x + 2x\cos(2x)}{\sin(2x)\cos(2x)} \) as x approaches 0.
We should write this expression as the sum of two terms.
\(\frac{2x}{\sin(2x)\cos(2x)} + {2x}{\sin(2x)}\)
Now rewrite the first fraction as the product of two fractions and then factor the result.
\(\frac{2x}{\sin(2x)} * \frac{1}{\cos(2x)} + \frac{2x}{\sin(2x)}\)
\(= \frac{2x}{\sin(2x)} \left(\frac{1}{\cos(2x)} + 1\right) \)
Multiply this out and then simplify. I did not show all the Algebra steps there because it is a pain but if you are in calculus this should be trivial for you.
You will this expression: \(\frac{1}{1} \left(\frac{1}{1} +1\right)\)
\(=2\)
Problem 6
Use the relation \(lim_{\theta \to 0} \frac{\sin \theta}{\theta} =1\) to find out the limit of the function below.
\(f(\theta)=\frac{\sin \theta}{\sin (2\theta)}\)
Multiply \(\frac{\sin \theta}{\sin (2\theta)} by \frac{2\theta}{2\theta} \).
This equals : \(\frac{1}{2} * \frac{\sin \theta}{\theta} * \frac{2\theta}{\sin (2\theta)}\).
We now have: \(\frac{1}{2} * 1 * \frac{1}{1} \).
So you should get \(\frac{1}{2}\) as your answer.
Conclusion
After doing a few problems you should be able to see how useful one-sided limits can be to you. This will be one of your premier tools later on to help you solve problems. Limits often exhibit some wild behavior. That is why we must look at both sides of any particular limit. Sometimes they may be exactly equal on both sides but many times they will not be. Graphing will help when it gets more complicated so I will be using more graphs as we progress.
Solving Rates of Change
One of the easiest techniques to use is called the difference method. When you put a curve on a graph it will have an infinite number of points between its starting and ending points. If you take any two points on this curve you can draw a straight line to connect them. This straight line might be perfectly flat with respect to the x-axis, go up, or go down.
The line that we drew will have a certain length according to the points on the graph. Length will equal the number of points on the graph that it moves through then. The line will also have a certain amount of points it moves up or down. This value could be 0, positive, or negative. If you divide this change in y by the change in x you will get the slope of the curve in between those two points. This will be important to remember as we move towards limits eventually.
Problem 1
Find the slope of the curve at the point (1,5).
$$ y= 3x^2 -2x +4 $$
The curve looks like this.
We will use the difference formula to solve basic derivatives for now.
$$ \frac{3(x + \bigtriangleup x)^2 - 2(x + \bigtriangleup x) + 4 - (3x^2 -2x + 4)}{\bigtriangleup x} $$
Lets multiple out these terms.
$$ \frac{3x^2 + 6x\bigtriangleup x + 3(\bigtriangleup x)^2 -2x - 2\bigtriangleup x +4 -3x^2 + 2x - 4} {\bigtriangleup x} $$
We can simplify this a lot now.
$$ \frac{6x\bigtriangleup x + 3(\bigtriangleup x)^2 - 2\bigtriangleup x} {\bigtriangleup x} $$
Further reducing this give us:
$$ 6x + 3\bigtriangleup x - 2 $$
This gives us:
$$ 6x - 2 $$
Our point we are evaluating is (1,5). The x-value of this point is 1. Substitute 1 for x in what is left of our equation.
$$ 6(1) - 2 $$
$$ 4 $$ is our slope at this point and the rate of change on the curve we are given.
Problem 2
Find the instantaneous rates of change of :
$$ y = \frac{2x}{x + 1} $$
when x = 2.
First we need to find:
$$ \frac{\bigtriangleup y}{\bigtriangleup x} $$
This basically says to find the change in y over the change in x. However, there is more to solving this.
What we are really saying is this.
$$ \bigtriangleup y = f(x + \bigtriangleup x) - f(x) $$
The original problem tells us that:
$$ y = \frac{2x}{x + 1} $$
Remember that y = f(x) and we are trying to find the change in y. So the change in y is:
$$ f( x + \bigtriangleup x) = \frac{2(x + \bigtriangleup x)}{x + \bigtriangleup x + 1} $$
Now we substitute what we know into this so far.
$$ \bigtriangleup y = \frac{ 2x + 2\bigtriangleup x}{x + \bigtriangleup x + 1} - \frac{2x}{x + 1} $$
Lets start by combining the terms:
$$ \bigtriangleup y = \frac{(2x + 2 * \bigtriangleup x)(x + 1) - 2x (x + \bigtriangleup x + 1)}{(x + 1)(x + \bigtriangleup x + 1)} $$
This of course can be simplified. Look for the terms that cancel.
$$ \bigtriangleup y = \frac{2 * \bigtriangleup x}{(x \bigtriangleup x + 1)(x + 1)(\bigtriangleup x)} $$
We now get a much better looking version.
$$ \frac{2}{(x + \bigtriangleup x + 1)(x + 1)} $$
Since this limit is approaching 0 we will try substituting 0 for change the in x value and see what kind of answer we get.
$$ \frac{\bigtriangleup y}{\bigtriangleup x} = \frac{2}{(x + 1)^2} $$
Now substitute 2 for x and see if we get an answer that makes sense. It looks good so I am happy with that. You can see we are edging towars the definition of limits here.
$$ = \frac{2}{(2 + 1)^2} = \frac{2}{9} $$
Problem 3
Find the average rates of change of:
$$ y = f(x) = x^2 - 2 $$ between the points $$ x = 3 and x = 4 $$
The average rate of change is defined by:
$$ \frac{\bigtriangleup y}{\bigtriangleup x} and \bigtriangleup y = f(x + \bigtriangleup x) - f(x) $$
We are given x = 3 and change in x = 4 - 3 = 1. So this gives us:
$$ y = f(x) = f(3) = 3^2 - 2 = 7 $$
The other point we are given is x = 4 so we evaluate for that as well.
$$ = y + \bigtriangleup y = f(x + \bigtriangleup x) = 4^2 - 2 =14 $$
This will yield:
$$ = \bigtriangleup y = f(x + \bigtriangleup x) - f(x) = f(4) - f(3) $$
Which reduces to:
$$ = (4^2 - 2) - (3^2 - 2) = 14 - 7 = 7 $$
The average rate of change of this function is 7.
Problem 4
Find the average rates of change for:
$$ y = \frac{1}{x} $$
The average rate of change is defined as:
$$ \frac{\bigtriangleup y}{\bigtriangleup x} with \bigtriangleup y = f(x + \bigtriangleup x) - f(x) $$
Since:
$$ f(x) = \frac{1}{x} and f(x + \bigtriangleup x) = \frac{1}{x + \bigtriangleup x} $$
And:
$$ \bigtriangleup y = \frac{1}{x + \bigtriangleup x} - \frac{1}{x} = \frac{x - (x + \bigtriangleup x)}{x(x + \bigtriangleup x)} $$
This will evaluate to:
$$ = \frac{- \bigtriangleup x}{x(x + \bigtriangleup x)} $$
This reduces to:
$$ \frac{\bigtriangleup y}{\bigtriangleup x} = \frac{- \bigtriangleup x}{x(x + \bigtriangleup x) \bigtriangleup x} = - \frac{1}{x(x + \bigtriangleup x)} $$
So that makes the average rate of change of this particular function as:
$$ \frac{-1}{x(x + \bigtriangleup x)} $$
Problem 5
Evaluate the rates of change in y=3x + 7 between x = -1 and x = 1.
$$ \bigtriangleup y = f(x + \bigtriangleup x) - f(x) $$
We will start with x = -1 and evaluate.
$$ \bigtriangleup x = 1 - (-1) = 2. $$
The change in x is equal to 2.
$$ \bigtriangleup y = f(x + \bigtriangleup x) - f(x) $$
Since we know the change in x we can now use it in this part of the equation.
$$ f(-1 + 2) - f(-1) $$
As you can see I just substituted the values.
$$ (3(1) + 7) - (3(-1) + 7) = 10 - 4 = 6 $$
So 6 is our answer as to the change in y.
Problem 6
Evaluate the rate of change in Y with respect to X at the point x = 5 when $$ 2y = x^2 + 3x - 1 $$
The rate of change is defined as $$ \bigtriangleup y = f(x + \bigtriangleup x) - f(x) $$
So, we do our inputs into the original equation.
$$ 2\bigtriangleup y = (x + \bigtriangleup x)^2 + 3(x + \bigtriangleup x) - 1 - (x^2 + 3x - 1) $$
As you can see this is a little more difficult version of what we have been working with. Pay attention to the sings and think it through and you will be fine.
We now have:
$$ x^2 + 2x * \bigtriangleup x + (\bigtriangleup x)^2 + 3x + 3\bigtriangleup x - 1 - x^2 - 3x + 1 $$
Theres a lot going on there but its not too bad.
Simplify things now.
$$ 2x * \bigtriangleup x + (\bigtriangleup x)^2 + 3\bigtriangleup x $$
Now we need to divide.
$$ \frac{2 \bigtriangleup y}{\bigtriangleup x} = \frac{2x * \bigtriangleup x}{\bigtriangleup x} + \frac{(\bigtriangleup x)^2}{\bigtriangleup x} + \frac{3 \bigtriangleup x}{\bigtriangleup x} = 2x + \bigtriangleup x + 3 $$
A lot of that will simplify down thank goodness. Here is what we have left after simplifying.
$$ \frac{\bigtriangleup y}{\bigtriangleup x} = x + \frac{\bigtriangleup x}{2} + \frac{3}{2} $$
Now we have:
$$ \frac{\bigtriangleup y}{\bigtriangleup x} = x + \frac{\bigtriangleup x}{2} + \frac{3}{2} = x + \frac{3}{2} $$
We are almost done now. If x = 5 then:
$$ \frac{\bigtriangleup y}{\bigtriangleup x} = 5 + \frac{3}{2} = 6 \frac{1}{2} $$
That is our rate of change of this curve at x = 5.
Problem 7
Find the derivative of the function:
$$ y = 2x^2 + 3x $$
The definition of of this type of derivative is:
$$ y'(x) = \frac{f(x + \bigtriangleup x) - f(x)}{\bigtriangleup x} $$
Does that definition look familiar? It should. It is what we have been using all along. What we have been doing all along is finding derivatives at points on a graph.
We have been given $$ f(x) = 2x^2 + 3x $$
We use our inputs now into the function.
$$ f(x + \bigtriangleup x) = 2(x + \bigtriangleup x)^2 + 3(x + \bigtriangleup x) $$
Now we start substituting.
$$ y'(x) = \frac{2(x + \bigtriangleup x)^2 + 3(x + \bigtriangleup x) - (2x^2 + 3x)}{\bigtriangleup x} $$
Lets reduce this down some.
$$ y'(x) = \frac{4x\bigtriangleup x + 2(\bigtriangleup x)^2 + 3\bigtriangleup x}{\bigtriangleup x} $$
Keep simplifying.
$$ y'(x) = 4x + 2\bigtriangleup x + 3 $$
The middle term drops because we assume it is approaching 0. We are then left with:
$$ y'(x) = 4x + 3 $$ as our derivative.