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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.

So:

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

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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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