# 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:
Recall:
L'Hôpital's Rule
Suppose that $\lim _{x\rightarrow \infty }f(x)$ and $\lim _{x\rightarrow \infty }g(x)$ are both zero or both $\pm \infty .$ If $\lim _{x\rightarrow \infty }{\frac {f'(x)}{g'(x)}}$ is finite or  $\pm \infty ,$ then $\lim _{x\rightarrow \infty }{\frac {f(x)}{g(x)}}\,=\,\lim _{x\rightarrow \infty }{\frac {f'(x)}{g'(x)}}.$ 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
$\lim _{x\rightarrow -3}{\frac {x^{3}-9x}{6+2x}}\,=\,{\frac {(-3)(-3-3)}{2}}\,=\,{\frac {18}{2}}\,=\,9.$ (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
$\lim _{x\rightarrow -\infty }{\frac {3x}{\sqrt {4x^{2}+x+5}}}\,=\,\lim _{x\rightarrow -\infty }{\frac {-3}{\sqrt {4+{\frac {1}{x}}+{\frac {5}{x^{2}}}}}}.$ So, we have
$\lim _{x\rightarrow -\infty }{\frac {3x}{\sqrt {4x^{2}+x+5}}}={\frac {-3}{\sqrt {4}}}={\frac {-3}{2}}.$ (a)$9$ (b)$+\infty$ (c)$-{\frac {3}{2}}$ 