# 009A Sample Final A, Problem 1

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

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

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

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

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

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,
${\displaystyle {\frac {1}{\infty }}=0,}$
and
${\displaystyle \lim _{x\rightarrow 0^{\pm }}{\frac {1}{x}}\,=\,\pm \infty .}$
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

${\displaystyle {\frac {0}{0}}}$   or   ${\displaystyle {\frac {\pm \infty }{\pm \infty }}.}$

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, ${\displaystyle x-1}$ can be factored as ${\displaystyle \left({\sqrt {x}}-1\right)\left({\sqrt {x}}+1\right)}$ , or as ${\displaystyle \left({\sqrt[{3}]{x}}-1\right)\left(\left({\sqrt[{3}]{x}}\right)^{2}+{\sqrt[{3}]{x}}+1\right)}$ , both of which could result in a factor that cancels in our fraction.
• We can apply l'Hôpital's Rule: Suppose ${\displaystyle c}$ is contained in some interval ${\displaystyle I}$. If ${\displaystyle \lim _{x\to c}f(x)=\lim _{x\to c}g(x)=0{\text{ or }}\pm \infty }$  and ${\displaystyle \lim _{x\to c}{\frac {f'(x)}{g'(x)}}}$   exists, and ${\displaystyle g'(x)\neq 0}$  for all ${\displaystyle x\neq c}$  in ${\displaystyle I}$, then ${\displaystyle \lim _{x\to c}{\frac {f(x)}{g(x)}}=\lim _{x\to c}{\frac {f'(x)}{g'(x)}}}$.
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 ${\displaystyle x}$ approaches 0. This is an indeterminate form, and we can apply l'Hôpital's Rule:
${\displaystyle \lim _{x\rightarrow 0}{\frac {\tan(3x)}{x^{3}}}\,\,{\overset {l'H}{=}}\,\,\lim _{x\rightarrow 0}{\frac {\sec ^{2}(3x)\cdot 3}{3x^{2}}}\,=\,\lim _{x\rightarrow 0}{\frac {3}{3x^{2}}}.}$
Now, ${\displaystyle x^{2}}$ can only be positive, so our limit can also only be positive. Thus, the limit is ${\displaystyle +\infty }$ .
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
${\displaystyle -\,{\frac {\sqrt {\frac {1}{x^{6}}}}{\frac {1}{x^{3}}}}\,=\,1,}$
since for negative values of ${\displaystyle x}$,
${\displaystyle {\sqrt {\frac {1}{x^{6}}}}\,=\,\left|{\frac {1}{x^{3}}}\right|\,=\,-\,{\frac {1}{x^{3}}}.}$
This means that
${\displaystyle \lim _{x\rightarrow -\infty }{\frac {\sqrt {x^{6}+6x^{2}+2}}{x^{3}+x-1}}\,=\,\lim _{x\rightarrow -\infty }{\frac {\sqrt {x^{6}+6x^{2}+2}}{x^{3}+x-1}}\cdot \left(-\,{\frac {\sqrt {\frac {1}{x^{6}}}}{\frac {1}{x^{3}}}}\right)}$
${\displaystyle =\,\lim _{x\rightarrow -\infty }-\,{\frac {\sqrt {1+{\frac {6x^{2}}{x^{6}}}+{\frac {2}{x^{6}}}}}{1+{\frac {x}{x^{3}}}+{\frac {1}{x}}}}}$
${\displaystyle =\,\lim _{x\rightarrow -\infty }-\,{\frac {\sqrt {1+{\frac {6}{x^{4}}}+{\frac {2}{x^{6}}}}}{1+{\frac {1}{x^{2}}}+{\frac {1}{x}}}}}$
${\displaystyle =\,-\,{\frac {\sqrt {1+0+0}}{1+0+0}}}$
${\displaystyle =\,-1.}$
Part (c):
Here, both the numerator and denominator go to zero as ${\displaystyle x}$ 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
${\displaystyle \lim _{x\rightarrow 3}{\frac {x-3}{{\sqrt {x+1}}-2}}\cdot {\frac {{\sqrt {x+1}}+2}{{\sqrt {x+1}}+2}}\,=\,\lim _{x\rightarrow 3}{\frac {(x-3)({\sqrt {x+1}}+2)}{\left({\sqrt {x+1}}\right)^{2}-\left(2\right)^{2}}}}$
${\displaystyle =\,\lim _{x\rightarrow 3}{\frac {(x-3)({\sqrt {x+1}}+2)}{x+1-4}}}$
${\displaystyle =\,\lim _{x\rightarrow 3}{\frac {(x-3)({\sqrt {x+1}}+2)}{x-3}}}$
${\displaystyle =\,\lim _{x\rightarrow 3}{\sqrt {x+1}}+2}$
${\displaystyle =\,4.}$
Alternatively, we can apply l'Hôpital's Rule:
${\displaystyle \lim _{x\rightarrow 3}{\frac {x-3}{{\sqrt {x+1}}-2}}\,\,{\overset {l'H}{=}}\,\,\lim _{x\rightarrow 3}{\frac {1}{\displaystyle {\frac {1}{2}}\cdot {\frac {1}{\sqrt {x+1}}}}}\,=\,{\frac {1}{{\frac {1}{2}}\cdot {\frac {1}{2}}}}\,=\,4.}$
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:
${\displaystyle \lim _{x\rightarrow 3}{\frac {x-1}{{\sqrt {x+1}}-1}}\,=\,{\frac {3-1}{{\sqrt {3+1}}-1}}\,=\,2.}$
Part (e):
Here, we again multiply by a fraction equal to one, noticing that for all ${\displaystyle x}$,
${\displaystyle {\frac {\frac {1}{x^{2}}}{\frac {1}{x^{2}}}}\,=\,1.}$
This means that
${\displaystyle \lim _{x\rightarrow \infty }{\frac {5x^{2}-2x+3}{1-3x^{2}}}\,=\,\lim _{x\rightarrow \infty }{\frac {5x^{2}-2x+3}{1-3x^{2}}}\cdot {\frac {\frac {1}{x^{2}}}{\frac {1}{x^{2}}}}\,=\,\lim _{x\rightarrow \infty }{\frac {5-{\frac {2}{x}}+{\frac {3}{x^{2}}}}{{\frac {1}{x^{2}}}-3}}\,=\,{\frac {5-0+0}{0-3}}\,=\,-\,{\frac {5}{3}}.}$