# 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

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