Limits, One-sided Limits, and a Handwritten Assignment

Hello Calculus students!

For this (partial) week, you should be re-reading sections 1.1-1.2 and reading sections 1.3 and (at least parts of) 1.4.

In addition to working on your 5 question assignment via WeBWorK (Assignment 2 – The Clone Wars), this Friday you will also need to turn in a handwritten assignment.  This will consist of the following problem (that was first described in class):

Handwritten Assignment 1

1(a) Consider the function g(x) = |x|.  Use one-sided limits to conclude that the instantaneous velocity of g(x) does not exist at x = 0.

1(b) Draw a graph of the function g(x) = |x| and explain (in a sentence or two) how the graph corroborates your work in part (a).

1(c) Explain why the function g(x) = |x| is continuous at x=0 and, in fact, why it is continuous at every x value.

I also want to use this blog post to include two more examples of (algebraically) computing some limits.

Example (1).  Let’s consider the function

f(x) = \frac{1}{\sqrt{x}-3} - \frac{6}{x-9}

This function has as its domain all x values where x \geq 0 and x \neq 9.  If we want to compute the limit \displaystyle \lim_{x\to9} f(x) we cannot just plug in f(9) since this produces an expression with zero in the denominator.  Instead, we must first algebraically manipulate the formula.  Taking a hint from today’s class we can multiply the top and bottom of the first fraction by \sqrt{x}+3.  In fact, we can also factor the denominator for the second expression:

f(x) = \frac{1}{\sqrt{x}-3} - \frac{6}{x-9} = \frac{1}{\sqrt{x}-3} - \frac{6}{(\sqrt{x}+3)(\sqrt{x}-3)} = \frac{\sqrt{x}+3}{(\sqrt{x}+3)(\sqrt{x}-3)} - \frac{6}{(\sqrt{x}+3)(\sqrt{x}-3)}

Now that this expression has a common denominator, we can write

f(x) = \frac{\sqrt{x}+3-6}{(\sqrt{x}+3)(\sqrt{x}-3)} = \frac{\sqrt{x}-3}{(\sqrt{x}+3)(\sqrt{x}-3)} = \frac{1}{\sqrt{x}+3}.

We should be careful and note that the cancellation used above works provided x \neq 9.  Since we are now taking a limit as x \to 9, we are allowed to do exactly this kind of simplification.  We find

\displaystyle \lim_{x\to 9} f(x) = \lim_{x\to 9} \frac{1}{\sqrt{x}+3} = \frac{1}{6}

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