# 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. In addition to this, we must consider that as a limit,
${\frac {1}{\infty }}=0,$ and
$\lim _{x\rightarrow 0^{\pm }}{\frac {1}{x}}\,=\,\pm \infty .$ In the latter case, the sign matters. 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, there 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.

Solution:

Part (a):
Note that both the numerator and denominator are continuous functions, and that the limit of each is 0 as $x$ approaches 0. This is an indeterminate form, and we can apply l'Hôpital's Rule:
$\lim _{x\rightarrow 0}{\frac {\tan(3x)}{x^{3}}}\,\,{\overset {l'H}{=}}\,\,\lim _{x\rightarrow 0}{\frac {\sec ^{2}(3x)\cdot 3}{3x^{2}}}\,=\,\lim _{x\rightarrow 0}{\frac {3}{3x^{2}}}.$ Now, $x^{2}$ can only be positive, so our limit can also only be positive. Thus, the limit is $+\infty$ .
Part (b):
In the case of limits at infinity, we can apply one other method. We can multiply our original argument by a fraction equal to one, and then can evaluate each term separately. Since we only need to consider values which are negative, we have that
$-\,{\frac {\sqrt {\frac {1}{x^{6}}}}{\frac {1}{x^{3}}}}\,=\,1,$ since for negative values of $x$ ,
${\sqrt {\frac {1}{x^{6}}}}\,=\,\left|{\frac {1}{x^{3}}}\right|\,=\,-\,{\frac {1}{x^{3}}}.$ This means that
$\lim _{x\rightarrow -\infty }{\frac {\sqrt {x^{6}+6x^{2}+2}}{x^{3}+x-1}}\,=\,\lim _{x\rightarrow -\infty }{\frac {\sqrt {x^{6}+6x^{2}+2}}{x^{3}+x-1}}\cdot \left(-\,{\frac {\sqrt {\frac {1}{x^{6}}}}{\frac {1}{x^{3}}}}\right)$ $=\,\lim _{x\rightarrow -\infty }-\,{\frac {\sqrt {1+{\frac {6x^{2}}{x^{6}}}+{\frac {2}{x^{6}}}}}{1+{\frac {x}{x^{3}}}+{\frac {1}{x}}}}$ $=\,\lim _{x\rightarrow -\infty }-\,{\frac {\sqrt {1+{\frac {6}{x^{4}}}+{\frac {2}{x^{6}}}}}{1+{\frac {1}{x^{2}}}+{\frac {1}{x}}}}$ $=\,-\,{\frac {\sqrt {1+0+0}}{1+0+0}}$ $=\,-1.$ Part (c):
Here, both the numerator and denominator go to zero as $x$ goes to 3, so we have an indeterminate form. We can choose to either apply l'Hôpital's Rule, or use the conjugate of the denominator. Using the conjugate, we find
$\lim _{x\rightarrow 3}{\frac {x-3}{{\sqrt {x+1}}-2}}\cdot {\frac {{\sqrt {x+1}}+2}{{\sqrt {x+1}}+2}}\,=\,\lim _{x\rightarrow 3}{\frac {(x-3)({\sqrt {x+1}}+2)}{\left({\sqrt {x+1}}\right)^{2}-\left(2\right)^{2}}}$ $=\,\lim _{x\rightarrow 3}{\frac {(x-3)({\sqrt {x+1}}+2)}{x+1-4}}$ $=\,\lim _{x\rightarrow 3}{\frac {(x-3)({\sqrt {x+1}}+2)}{x-3}}$ $=\,\lim _{x\rightarrow 3}{\sqrt {x+1}}+2$ $=\,4.$ Alternatively, we can apply l'Hôpital's Rule:
$\lim _{x\rightarrow 3}{\frac {x-3}{{\sqrt {x+1}}-2}}\,\,{\overset {l'H}{=}}\,\,\lim _{x\rightarrow 3}{\frac {1}{\frac {1}{2}}\cdot {\frac {1}{\sqrt {x+1}}}}}\,=\,{\frac {1}{{\frac {1}{2}}\cdot {\frac {1}{2}}}}\,=\,4.$ Part (d):
This problem is meant to confuse you. It looks like you should multiply by a conjugate, but instead you can just plug in and evaluate:
$\lim _{x\rightarrow 3}{\frac {x-1}{{\sqrt {x+1}}-1}}\,=\,{\frac {3-1}{{\sqrt {3+1}}-1}}\,=\,2.$ Part (e):
Here, we again multiply by a fraction equal to one, noticing that for all $x$ ,
${\frac {\frac {1}{x^{2}}}{\frac {1}{x^{2}}}}\,=\,1.$ This means that
$\lim _{x\rightarrow \infty }{\frac {5x^{2}-2x+3}{1-3x^{2}}}\,=\,\lim _{x\rightarrow \infty }{\frac {5x^{2}-2x+3}{1-3x^{2}}}\cdot {\frac {\frac {1}{x^{2}}}{\frac {1}{x^{2}}}}\,=\,\lim _{x\rightarrow \infty }{\frac {5-{\frac {2}{x}}+{\frac {3}{x^{2}}}}{{\frac {1}{x^{2}}}-3}}\,=\,{\frac {5-0+0}{0-3}}\,=\,-\,{\frac {5}{3}}.$ 