009A Sample Final A, Problem 1

1. Find the following limits:
   (a)   $\lim _{x\rightarrow 0}{\frac {\tan(3x)}{x^{3}}}.$

   (b) $\lim _{x\rightarrow -\infty }{\frac {\sqrt {x^{6}+6x^{2}+2}}{x^{3}+x-1}}.$

   (c)   $\lim _{x\rightarrow 3}{\frac {x-3}{{\sqrt {x+1}}-2}}.$

   (d)   $\lim _{x\rightarrow 3}{\frac {x-1}{{\sqrt {x+1}}-1}}.$

   (e)  $\lim _{x\rightarrow \infty }{\frac {5x^{2}-2x+3}{1-3x^{2}}}.$

Foundations:
When evaluating limits of rational functions, the first idea to try is to simply plug in the limit. Unfortunately, most (but not all) exam questions require more work. Many of them will evaluate to an indeterminate form, or something of the form
${\frac {0}{0}}$   or   ${\frac {\pm \infty }{\pm \infty }}.$
In this case, here are several approaches to try:
We can multiply the numerator and denominator by the conjugate of the denominator. This frequently results in a term that cancels, allowing us to then just plug in our limit value. We can factor a term creatively. For example, $x-1$ can be factored as $\left({\sqrt {x}}-1\right)\left({\sqrt {x}}+1\right)$ , or as $\left({\sqrt[{3}]{x}}-1\right)\left(\left({\sqrt[{3}]{x}}\right)^{2}+{\sqrt[{3}]{x}}+1\right)$ , both of which could result in a factor that cancels in our fraction. We can apply l'Hôpital's Rule: Suppose $c$ is contained in some interval $I$. If $\lim _{x\to c}f(x)=\lim _{x\to c}g(x)=0{\text{ or }}\pm \infty$  and $\lim _{x\to c}{\frac {f'(x)}{g'(x)}}$   exists, and $g'(x)\neq 0$  for all $x\neq c$  in $I$, then $\lim _{x\to c}{\frac {f(x)}{g(x)}}=\lim _{x\to c}{\frac {f'(x)}{g'(x)}}$.
Note that the first requirement in l'Hôpital's Rule is that the fraction must be an indeterminate form. This should be shown in your answer for any exam question.

Part (a):

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