# 009A Sample Final 1, Problem 1

In each part, compute the limit. If the limit is infinite, be sure to specify positive or negative infinity.

(a)   $\lim _{x\rightarrow -3}{\frac {x^{3}-9x}{6+2x}}$ (b)   $\lim _{x\rightarrow 0^{+}}{\frac {\sin(2x)}{x^{2}}}$ (c)   $\lim _{x\rightarrow -\infty }{\frac {3x}{\sqrt {4x^{2}+x+5}}}$ Foundations:
L'Hôpital's Rule, Part 1

Let  $\lim _{x\rightarrow c}f(x)=0$ and  $\lim _{x\rightarrow c}g(x)=0,$ where  $f$ and  $g$ are differentiable functions

on an open interval  $I$ containing  $c,$ and  $g'(x)\neq 0$ on  $I$ except possibly at  $c.$ Then,   $\lim _{x\rightarrow c}{\frac {f(x)}{g(x)}}=\lim _{x\rightarrow c}{\frac {f'(x)}{g'(x)}}.$ Solution:

(a)

Step 1:
We begin by factoring the numerator. We have

$\lim _{x\rightarrow -3}{\frac {x^{3}-9x}{6+2x}}\,=\,\lim _{x\rightarrow -3}{\frac {x(x-3)(x+3)}{2(x+3)}}.$ So, we can cancel  $x+3$ in the numerator and denominator. Thus, we have

$\lim _{x\rightarrow -3}{\frac {x^{3}-9x}{6+2x}}\,=\,\lim _{x\rightarrow -3}{\frac {x(x-3)}{2}}.$ Step 2:
Now, we can just plug in  $x=-3$ to get

${\begin{array}{rcl}\displaystyle {\lim _{x\rightarrow -3}{\frac {x^{3}-9x}{6+2x}}}&=&\displaystyle {\frac {(-3)(-3-3)}{2}}\\&&\\&=&\displaystyle {\frac {18}{2}}\\&&\\&=&\displaystyle {9.}\end{array}}$ (b)

Step 1:
We proceed using L'Hôpital's Rule. So, we have

${\begin{array}{rcl}\displaystyle {\lim _{x\rightarrow 0^{+}}{\frac {\sin(2x)}{x^{2}}}}&=&\displaystyle {\lim _{x\rightarrow 0^{+}}{\frac {2\cos(2x)}{2x}}}\\&&\\&=&\displaystyle {\lim _{x\rightarrow 0^{+}}{\frac {\cos(2x)}{x}}.}\\\end{array}}$ Step 2:
This limit is   $\infty .$ (c)

Step 1:
We have

$\lim _{x\rightarrow -\infty }{\frac {3x}{\sqrt {4x^{2}+x+5}}}=\lim _{x\rightarrow -\infty }{\frac {3x}{{\sqrt {x^{2}(4+{\frac {1}{x}}+{\frac {5}{x^{2}}}}})}}.$ Since we are looking at the limit as  $x$ goes to negative infinity, we have  ${\sqrt {x^{2}}}=-x.$ So, we have

$\lim _{x\rightarrow -\infty }{\frac {3x}{\sqrt {4x^{2}+x+5}}}\,=\,\lim _{x\rightarrow -\infty }{\frac {3x}{-x{\sqrt {4+{\frac {1}{x}}+{\frac {5}{x^{2}}}}}}}.$ Step 2:
We simplify to get

${\begin{array}{rcl}\displaystyle {\lim _{x\rightarrow -\infty }{\frac {3x}{\sqrt {4x^{2}+x+5}}}}&=&\displaystyle {\lim _{x\rightarrow -\infty }{\frac {-3}{\sqrt {4+{\frac {1}{x}}+{\frac {5}{x^{2}}}}}}}\\&&\\&=&\displaystyle {-{\frac {3}{\sqrt {4}}}}\\&&\\&=&\displaystyle {-{\frac {3}{2}}.}\end{array}}$ (a)    $9$ (b)    $\infty$ (c)    $-{\frac {3}{2}}$ 