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Limits vs Values

Look at the function f(x) in orange below:


We are going to answer 4 questions about this graph.  They are all related to each other, but different questions.  Seeing the difference will help us sort out the difference between a function value and a limit.

Q1:  Find f(1)

Q2:  Find  limx→1 f(x)

Q3:  Find  limx→1+ f(x)

Q4:  Find  limx→1 f(x)

OK….. Try to answer these questions with what you know… Then continue reading to see the answers and explanations!

Continue reading Limits vs Values

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Basic Concept of a Limit

Here is a brief example on the concept of a limit:

Look at the function f(x) in orange below:


Question 1.  Find f(3)

Explanation of question 1: Find the value of the function when you plug in 3.  What is the height of the function at the exact moment when x=3?

Answer 1:  The function is undefined at x=3.  There is a hole when x=3.

So, f(3) is undefined.

Question 2:  Find limx→3f(x)

Explanation of question 2:  We are being asked to find what the function is doing around (but not at) 3.  What is happening to the path of the function on either side of 3?

In order to find limx→3f(x), we must confirm that limx→3+  f(x) and limx→3–  f(x) both exist and are equal to each other.

So, let’s find limx→3+  f(x).  What is happening to the function values as you approach x=3 from the right-hand side?  Literally run your finger along as if x=4, then x=3.5, then x=3.1.  What value is the function getting closer to?


The function is approaching a height of 4.

Let’s find limx→3–  f(x).  What is happening to the function values as you approach x=3 from the left-hand side?  Literally run your finger along as if x=1, then x=2, then x=2.9.  What value is the function getting closer to?


The function is also approaching a height of 4.


limx→3+  f(x) = 4

limx→3–  f(x) = 4

Since, the left-handed limit at 3 and right-handed limit at 3 exist and are equal, this gives:

limx→3f(x) = 4.

So, to summarize, here are 4 different things we found.  They are related, but not necessarily the same:

f(3) is undefined

limx→3+  f(x) = 4

limx→3–  f(x) = 4

limx→3f(x) = 4

Are you ready to try one on your own? Click here! (Don’t worry, I’ll walk you through the solutions too)

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Limit Example

Q:  What is limx→2 (x-2)/|x-2|

A:  This question is not too bad if you know what the function (x-2)/|x-2| looks like graphically.  But, let’s say you don’t.

We are going to “talk our way” through this problem to help solidify the concept of a limit.

If you plug in 2 to the function, you are finding the value of the function when x=2.  This is important, and related, though it is not the limit.  This is even sometimes a skill used to help us find the limit, but it is still not the limit.  Sometimes the function value is equal to the function limit, which can also be confusing, but not all the time.  Let’s find the value of the function when x=2:

(2-2)/|2-2| = 0/0 = undefined.

Okay.  This function is undefined when x=2.  This means there is a hole, or an asymptote, or a break or a jump or some disruption in the continuity of the function. Continue reading Limit Example

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One-sided Limit Example

Q:  Find the one-sided limit (if it exists):

limx→-1–     (x+1)/(x4-1)

A:  So we need to find the limit of this function (x+1)/(x4-1) as x approaches -1 from the left.  Remember, from the left means as x gets closer and closer to -1, but is still smaller.

The concept: What is happening to this function as x = -2, x = -1.5, x = -1.1, x = -1.0001, etc…

We test first and plug -1 into the function: (-1+1)/((-1)4-1) = 0/0

Whenever you get 0/0, that is your clue that maybe you need to do “more work” before just plugging in or jumping to conclusions.

So, let’s try “more work” — usually that means simplifying.  I see that the denominator can factor.  We have:

(x+1) / (x4-1) = (x+1) / [(x2-1)(x2+1)]

Let’s keep factoring the denominator:

(x+1) / [(x-1)(x+1)(x2+1)]

Now, it appears there is a “removable hole” in the function.  This means, we can remove this hole by reducing the matching term in the numerator with the matching term in the denominator:

(x+1) / [(x-1)(x+1)(x2+1)]

= 1 / [(x-1)(x2+1)]

Notice that hole exists when x = -1 (and it was removable! This is good news for us since we are concerned with the nature of the function as x approaches -1)

Continue reading One-sided Limit Example

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Integration: Sines, Cosines and U-Substitution

Q:  ∫cos5(x)sin4(x)dx

A: Some books have tables and charts to memorize on how to integrate these types of problems: what to do if the power is odd on one but even on the other, etc (called reduction formulas)… Boring… Who has time, space or desire to memorize formulas? Let’s solve the damn problem.

First, I see that there are sines and cosines, and we know that one is the derivative of the other (more or less).  This tells me that u-substitution is likely going to come up.

Remember: whenever you see a function and its derivative present in a problem, you want to be thinking u-substitution!

I also know that by using the Pythagorean Identity, sin2(x)+cos2(x)=1, I can convert an “even number” of cosines to sines and vice versa.

So, now that I know u-substitution is most likely, I want to leave behind one function to be “du” and the rest should be converted to “u’s”.

Cosine is literally the “odd man out”.  There is an odd number of cosines, so I will leave one cosine behind to eventually serve as du and convert the rest like so:





Now, I can do a fairly clean u-substitution:

Let u = sin(x)

Then, du = cos(x)dx

Continue reading Integration: Sines, Cosines and U-Substitution

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Integration: Disk (Washer Method) vs Shell Method

Here is an example to help you understand and visualize the difference between the disk method and the shell method.  I will do the same problem twice: once using disk method and once using shell method.  Keep in mind: not all problems are equally as easily solved with both methods — that’s why we have multiple methods!  Some problems may be easy to do with the shell method but nearly impossible with the disk method, or vice versa.

If you haven’t read the blog posts that discuss the basic differences between the disk method and the shell method, read those first:

To get the most out of this problem, grab a pen and paper and do the problem along with me: draw each picture and write each equation.

Q:  Find the volume obtained by rotating the area contained by y = √(x), y = 0, and x = 4 around the y-axis.

A:  First, draw a picture of the area that is to be rotated.




Visualize the rotation.  Actually give the shape a name to help solidify it in your mind.  I think of this shape as a cinder cone:


Obviously this isn’t a perfect visual match, but giving a real image to an abstract shape really helps with mentally processing the steps.

Method 1:  Disk (washer) Method.  Remember, the disk and washer method are the same thing.  In the disk method, we visualize stacked circles (or pancakes, as I like to say).  The washer method is the same stacked pancakes with holes (like washers or donuts).

I like to complete every problem with the same thought process.  Follow these steps for all volume/integration problems and you will get the hang of it!

Step 1.  Determine if this is a dx or a dy problem.

Which way are we stacking our pancakes?


Pancakes are being stacked vertically, so this is a “dy problem”.  This means the limits of integration and the equations used will all be in terms of y.

?? ____?____  dy

Step 2.  Find the limits of integration.

Where does the pancake stacking start?  Where does it end?  Remember — in terms of y since this is a dy problem.

Pancakes start at y=0 and end at the place where y = √(x) and x = 4 intersect (the top of the cinder cone).  These bounds intersect at the point (4, 2).

So, pancake stacking starts at y=0 and end at y=2.

02 ____?____  dy

Step 3.  Find the equation of the areas that are being stacked.

We’ve pre-decided that we are going to use the disk (washer) method.  There are definitely washers happening, because there is a big outer circle minus a hole (to create the cinder cone).  So, we need to stack outer pancakes minus inner pancakes (big circles minus small circles):

Area = πR2 – πr2

Let’s start with the area of the big circles, which I’ve called πR2.  What is the radius, R, of the big circles?  Is it changing throughout the problem or is it constant?


The radius of the larger circles is constant.  The larger circles have a radius of 4 throughout the entire cinder cone:  R = 4

Now look at the smaller circles.  Is the radius, r, constant or changing?


Notice the radius of the smaller circles is changing.  This radius is a horizontal distance, starting from the y-axis and moving on out.  The radius is x: r = x.

BUT WAIT…. remember, this is a dy problem.  All equations need to be in terms of y!  So, using r = x will not help.  We need to find a way to represent x in terms of y.  Fortunately, we have the equation to help: y = √(x), so, x = y2

Good.  So, R = 4 and r = y2

Our equation is now complete and ready to solve:

02 π(4)2 – π(y2)2 dy

I will leave the solving to you… but as a final answer I get 128π/5 (approx = 80.425)

Now… Are we ready to solve this same problem using the shell method?

Method 2:  Shell method.  Instead of visualizing stacked pancakes to create our cinder cone, we will visualize stacked “Russian Dolls” (cylinders).  We are going to stack these cylinders so tightly together that the lateral area of the cylinders will stack to create volume.


Back to our steps:

Step 1.  Determine if this is a dx or a dy problem.

The dolls are being stacked inside of each other and on outward, expanding along the x-axis.  This is a “dx problem”.  This means the limits of integration and the equations used will all be in terms of x.

?? ____?____  dx

Step 2.  Find the limits of integration.

Where does the cylinder stacking start and where does it end?  In other words: what is the radius of our smallest cylinder and what is the radius of our largest cylinder?

The smallest cylinder has a radius of x=0 and the largest cylinder has a radius of x=4

04 ____?____  dx

Step 3.  Find the equation of the areas that are being stacked.

We are stacking lateral areas of a cylinder, which has equation: 2πrh (r is the radius of the cylinder, h is the height).

Let’s figure out the radius, r, of a random cylinder/doll in our shape.  The radius does change, so it is a variable.


But, it does not necessarily depend on the functions.  The radius is simply an x-value that continues to grow until we hit the wall of x = 4.

So, r = x.

Now, let’s figure out the height, h, of a random cylinder/doll in our shape.  The height is changing and it is definitely affected by the functions.  The height is the y-value of the bounding function: h = y…. BUT WAIT… remember, this is a dx problem — no y’s allowed. So, use the equation: y = √(x).  Therefore, h = √(x)

So, we have:

04  2π(x)(√(x)) dx

I will leave the solving to you… but as a final answer I get 128π/5 (approx = 80.425) — no coincidence that this is the same answer obtained by method 1!

Same shape, two methods, same answer.  Phew.


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When to use Disk Method versus Shell Method, Part 2

To start, read: When to use Disk Method versus Shell Method, Part 1 to get a general visual.  This is a little more detailed:

Volume should be thought of as infinitely stacked area. In the disk method, you are infinitely stacking circles (think pancakes). In the shell method, you are infinitely stacking lateral surface areas of cylinders (think “Russian Dolls” that stack inside of each other)

The Disk Method: Since you are stacking pancakes, the general formula that you will be integrating is π*r2. If the radius of each disk is changing throughout the shape, the radius, r, will be a function, dependent on either y or x, depending on how you are rotating.


The Shell Method: Since you are stacking lateral areas of cylinders, the general formula that you will be integrating is 2*π*r*h (lateral area of cylinder formula). The radius, r, will be a simple function involving an x or a y. The height, h, will depend on the functions that are being rotated.


Now that you have an understanding of the concept, view this example… I solve a volume problem using the disk method.  Then, I solve the exact same problem using the shell method:   Integration Example: Disk Method vs. Shell Method

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Finding the limit example

Q: Find the limit (as x approaches 3) of (x3 – 27) / (3-x)

A:  The first thing to do when finding a basic limit is try plugging in the number in question (3).

So, plug in 3 to get:

(33 – 27) / (3-3) = 0/0 <– if you get 0/0 or infinity/infinity that means there is more work to be done.  However, if you had just got a number like 4 or something, that would’ve been your answer!

OK, we got 0/0 so that means more work.  More work could mean many things (apply different rules, factor and cancel, simplify, etc).  In this case, it appears we can factor, so we try that:

(x3 – 27) / (3-x) = (x – 3)(x2 + 3x + 9)/ (3 – x)

Now, here comes some tricky insight.  I notice that the (x – 3) on top is very similar to the (3 – x) on the bottom.  I am going to factor a “-1” out of the (x – 3) that is on top.

Notice:  -1(3 – x) = (x – 3)

So, the numerator becomes:

-1(x – 3)(x2 + 3x + 9)/ (3 – x)

Now, the (x – 3) term cancels from the top and bottom to leave:

-1(x2 + 3x + 9)

So, we are trying to find the limit (as x approaches 3) of -1(x2 + 3x + 9).  We have “removed the hole” — the factor (3 – x) was a “removable hole” that was causing calculation problems.  In the simplified version, we can plug in the value 3 to now calculate where that hole was occuring:

Lim (x –> 3) of -1(x2 + 3x + 9) = -1(32 + 3(3) + 9) = -1(9 + 9 + 9) = -27

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Quotient Rule Example

Q:  Find dy/dx of

y = x / sqrt(x2 + 1)

A:  To find dy/dx (the derivative), we will need to use the quotient rule since we have a function over a function.  See? We are in the form:

y = f / g where f and g are two different functions of x.

In this form, the quotient rule tells us if:

y = f / g

y ‘ = (g * f ‘ – f * g ‘ ) / g2

So, we know:

f = x

g = sqrt(x2 + 1)

f ‘ = 1   <— basic derivative

g ‘ = (this takes more work…. First, rewrite “g”):

g = sqrt(x2 + 1) = (x2 + 1)1/2

Now, use a power rule and a chain rule to find g ‘ like so:

g ‘ = 1/2 (x2 + 1) – 1/2 (2x)

[that’s the derivative of the outside * the derivative of the inside]

Clean up g ‘:

g ‘ = x (x2 + 1) – 1/2

So now we have all of the players:  f, f ‘ , g, g ‘:

f = x

g = (x2 + 1)1/2

f ‘ = 1

g ‘ = x (x2 + 1) – 1/2

Now, we said:  y ‘ = (g * f ‘ – f * g ‘ ) / g2, so plug in the pieces then simplify like so:

y ‘ = [ (x2 + 1)1/2(1) – (x)(x (x2 + 1) – 1/2 ) ] / [(x2 + 1)1/2]2


y ‘ = [ (x2 + 1)1/2 – x2(x2 + 1) – 1/2  ] / (x2 + 1)

Multiply top and bottom by (x2 + 1)1/2

[ (x2 + 1)1/2(x2 + 1)1/2 (x2 + 1)1/2x2(x2 + 1) – 1/2  ] / (x2 + 1)1/2(x2 + 1) =

[ (x2 + 1)- x2 ] / (x2 + 1)3/2

1 / (x2 + 1)3/2


So, if:

y = x / sqrt(x2 + 1)

dy/dx = 1 / (x2 + 1)3/2

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Critical points, Max/Min, Points of Inflection, Concavity – all in one!

Q:  Consider the function:  f(x) = (1/3) x3 – x2 – 15x + 1

Find / tell me about:

(a) critical points

(b) the local minimum / maximums [relative extrema]

(c) points of inflection

(d) concavity

A:  OK.  Get out your pen and paper and be writing with me as I calculate and explain.  This is the best way for you to learn.

(a)  The critical points are where the derivative equals 0 or is undefined.  So, take the derivative to start:

f(x) = (1/3) x3 – x2 – 15x + 1

f ‘ (x) = x2 – 2x- 15  <– you should get this!

Now, we need to find where the derivative = 0 (or is undefined — since the derivative is a quadratic, it is never undefined, so we don’t have to worry about that part).

0 = x2 – 2x- 15

Solve by factoring:

0 = (x – 5)(x + 3)

So, the critical points are when x = 5, -3.

(b)  To determine if the critical points are local maximum or minimums, we need to do the second derivative test.

Recap on second derivative test:  Plug a critical point into the second derivative. 

*If the end result is positive, this means the shape is concave up (smiley face) which means we have a minimum.

*If the end result is negative, this means the shape is concave down (frowny face) which means we have a maximum.

So, take the second derivative:

f(x) = (1/3) x3 – x2 – 15x + 1  [original function]

f ‘ (x) = x2 – 2x- 15  [1st derivative]

f ” (x) = 2x – 2  [2nd derivative]

Now, we have critical points of x = 5 and x = -3.  Deal with them one at a time in the second derivative to perform the second derivative test.

f ” (5) = 2(5) – 2 = 10 – 2 = 8 <– this is positive, which means when x = 5 we have a local minimum!

f ” (-3) = 2(-3) – 2 = -6 – 2 = -8 <– this is negative, which means when x = -3 we have a local maximum!

If you wanted to find the (x, y) point of the actual max and min, you would plug the x-values into the original equations to find the corresponding y values:

f(5) = (1/3) (5)3 – (5)2 – 15(5) + 1 = -172/3

So, f(x) has a local minimum at the point (5, -172/3)

f(-3) = (1/3) (-3)3 – (-3)2 – 15(-3) + 1 = 28

So, f(x) has a local maximum at the point (-3, 28)

(c)  The points of inflection are where the concavity of the function changes from concave up to concave down (and vice versa).  Our POI candidates can be found by setting the second derivative = 0 and finding the x candidates:

f ” (x) = 2x – 2

0 = 2x – 2 and solve for x:

x = 1

So, x = 1 is a candidate for a point of inflection (POI) — it is not a guaranteed POI though.

We need to interval test into the second derivative to see what is happening (generally, you would need to test in between every x value.  We only have 1 x value, so we just need to interval test on either side of it):

Plot your x values on a number line.  In our case, the only value is x = 1.  Pick a number on the left side of 1.  Let’s pick an easy number, 0:

Plug 0 into the second derivative:

f ” (0) = 2(0) – 2 = -2 <– since this is negative:  this interval, to the left of x = 1, is concave down

Now, pick a number to the right of x = 1.  Let’s pick 2:

f ” (2) = 2(2) – 2 = 2 <– since this is positive:  this interval, to the right of x = 1, is concave up

Therefore, when x = 1 we do have a point of inflection (since we changed from concave down to concave up).

If you want the (x, y) point, plug the x value into the original equation to find y:

f(1) = (1/3) (1)3 – (1)2 – 15(1) + 1 = -44/3

There is a point of inflection at (1, -44/3).

(d)  Now to analyze concavity, which we have basically already done in part (c)!

To the left of x = 1, we determined the graph to be concave down.

To the right of x = 1, we determined the graph to be concave up.

Concave down on the interval (-infinity, 1)

Concave up on the interval (1, infinity)
We have done it.  That was a lot to do.  Enjoy.

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Rolle’s Theorem: Concept and Example

Q:  Can you explain Rolle’s Theorem?

A:  Rolle’s Theorem is a specific application of the Mean Value Theorem.

What you need to apply Rolle’s Theorem:  a function that is differentiable everywhere on the interval you are looking at.  Basically, you need a smooth function (rolling hills, no points, breaks, sharp corners, holes, etc…)

Rolle’s says that if the function attains the same height at two different points (and meets the “smooth function” criteria) then there must be a point in the middle somewhere where the derivative of the function is zero.


Consider the function (I just made it up for example sake):
f(x) = x2 – 8x

Now, I know:

*This is a smooth / differentiable function

*And, f(2) = -12 and f(6) = -12  [just plug in 2 and then plug in 6 to verify… 2 and 6 are numbers I picked for the sake of the example]

So, Rolle’s Thm tells me that since f(2) and f(6) both = -12 and the function is differentiable…. There is an x value somewhere between 2 and 6 where the derivative of the function is 0.  Essentially, there is a “turn around point”  In order to go from the point (2, -12) to the point (6, -12), you must eventually “turn around” <— which gives you a derivative of 0.

That is all Rolle’s tells you.  It doesn’t tell you what the point is (and it doesn’t tell you if there is just 1 point where this happens or if there are more).  Rolle’s just tells you there is at least 1 point in that interval where the derivative will be 0.  You have to do the work to find that point (or points).

Let’s do it for fun:

f(x) = x2 – 8x

Take the derivative:

f ‘ (x) = 2x – 8

Set = 0 to find the turnaround point:

0 = 2x – 8

Solve for x (I won’t show those steps):

x = 4.

So, the derivative is 0 when x = 4.  And of course, x = 4 is definitely on the interval from 2 to 6 as Rolle’s predicted.

***Keep in mind, this is a very basic example.  It turned out the the derivative only had 1 zero anyway.  The derivative could have had MANY zeroes…. Rolle’s just would have guaranteed us that one of them (or more) was on our interval!

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Linearization: Concept and Example

Q:  Find the linearization of f(x) = ex at x = 0

A:  First, some concept:  Linearization is the act of finding a “linear function” that can approximate the given function on or around a given point.  In this problem, we want to find a line that models the shape of ex when you are around the point x = 0.

Step 1:  Find an (x, y) point on the function in question.

The function is f(x) = ex .  The x part of the point is 0.  Plug that in to find y:

f(0) = e0 = 1

So, the point is (0, 1).  Hold this point.  We will need it for later.

Essentially, we want to find a line that follows that patterns of f(x) = ex and goes through the point (0, 1).

So, to model the pattern, we need to slope of f(x) = ex at the given point.

Step 2:  Find the derivative (slope) of the function at the given point:

f(x) = ex

Find the derivative:

f ‘ (x) = ex

Now, find the slope at (0, 1):

f ‘ (0) = e0 = 1

So, the slope is 1.  m = 1

Step 3:  Find a line that has the same slope as the function that goes through the given point:

We need a line with slope of 1 that goes through the point (0, 1).

Start with your equation of a line:

y = mx + b

We know m = 1, so:

y = x + b

Plug in the (x, y) point to find “b”

1 = 0 + b

1 = b

So, the equation of the line is y = x + 1

In summation, to approximate values of f(x) = ex around where x = 0 you can use the line y = x + 1