# Difference between revisions of "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) ${\displaystyle \lim _{x\rightarrow -3}{\frac {x^{3}-9x}{6+2x}}}$
b) ${\displaystyle \lim _{x\rightarrow 0^{+}}{\frac {\sin(2x)}{x^{2}}}}$
c) ${\displaystyle \lim _{x\rightarrow -\infty }{\frac {3x}{\sqrt {4x^{2}+x+5}}}}$
Foundations:
Recall:
L'Hôpital's Rule
Suppose that ${\displaystyle \lim _{x\rightarrow \infty }f(x)}$  and ${\displaystyle \lim _{x\rightarrow \infty }g(x)}$  are both zero or both ${\displaystyle \pm \infty .}$
If ${\displaystyle \lim _{x\rightarrow \infty }{\frac {f'(x)}{g'(x)}}}$  is finite or  ${\displaystyle \pm \infty ,}$
then ${\displaystyle \lim _{x\rightarrow \infty }{\frac {f(x)}{g(x)}}\,=\,\lim _{x\rightarrow \infty }{\frac {f'(x)}{g'(x)}}.}$

Solution:

(a)

Step 1:
We begin by factoring the numerator. We have
${\displaystyle \lim _{x\rightarrow -3}{\frac {x^{3}-9x}{6+2x}}\,=\,\lim _{x\rightarrow -3}{\frac {x(x-3)(x+3)}{2(x+3)}}.}$
So, we can cancel ${\displaystyle x+3}$  in the numerator and denominator. Thus, we have
${\displaystyle \lim _{x\rightarrow -3}{\frac {x^{3}-9x}{6+2x}}\,=\,\lim _{x\rightarrow -3}{\frac {x(x-3)}{2}}.}$
Step 2:
Now, we can just plug in ${\displaystyle x=-3}$  to get
${\displaystyle \lim _{x\rightarrow -3}{\frac {x^{3}-9x}{6+2x}}\,=\,{\frac {(-3)(-3-3)}{2}}\,=\,{\frac {18}{2}}\,=\,9.}$

(b)

Step 1:
We proceed using L'Hôpital's Rule. So, we have
${\displaystyle {\begin{array}{rcl}\displaystyle {\lim _{x\rightarrow 0^{+}}{\frac {\sin(2x)}{x^{2}}}}&=&\displaystyle {\lim _{x\rightarrow 0^{+}}{\frac {2\cos(2x)}{2x}}}\\&&\\&=&\displaystyle {\lim _{x\rightarrow 0^{+}}{\frac {\cos(2x)}{x}}.}\\\end{array}}}$
Step 2:
This limit is  ${\displaystyle +\infty .}$

(c)

Step 1:
We have
${\displaystyle \lim _{x\rightarrow -\infty }{\frac {3x}{\sqrt {4x^{2}+x+5}}}=\lim _{x\rightarrow -\infty }{\frac {3x}{{\sqrt {x^{2}(4+{\frac {1}{x}}+{\frac {5}{x^{2}}}}})}}.}$
Since we are looking at the limit as ${\displaystyle x}$ goes to negative infinity, we have ${\displaystyle {\sqrt {x^{2}}}=-x.}$
So, we have
${\displaystyle \lim _{x\rightarrow -\infty }{\frac {3x}{\sqrt {4x^{2}+x+5}}}\,=\,\lim _{x\rightarrow -\infty }{\frac {3x}{-x{\sqrt {4+{\frac {1}{x}}+{\frac {5}{x^{2}}}}}}}.}$
Step 2:
We simplify to get
${\displaystyle \lim _{x\rightarrow -\infty }{\frac {3x}{\sqrt {4x^{2}+x+5}}}\,=\,\lim _{x\rightarrow -\infty }{\frac {-3}{\sqrt {4+{\frac {1}{x}}+{\frac {5}{x^{2}}}}}}.}$
So, we have
${\displaystyle \lim _{x\rightarrow -\infty }{\frac {3x}{\sqrt {4x^{2}+x+5}}}={\frac {-3}{\sqrt {4}}}={\frac {-3}{2}}.}$
(a)${\displaystyle 9}$
(b)${\displaystyle +\infty }$
(c)${\displaystyle -{\frac {3}{2}}}$