# Difference between revisions of "009A Sample Final 1, Problem 1"

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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:
L'Hôpital's Rule, Part 1

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

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

Solution:

(a)

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

${\displaystyle \lim _{x\rightarrow -3}{\frac {x^{3}-9x}{6+2x}}\,=\,\lim _{x\rightarrow -3}{\frac {x(x-3)(x+3)}{2(x+3)}}.}$

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

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

(b)

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

${\displaystyle {\begin{array}{rcl}\displaystyle {\lim _{x\rightarrow 0^{+}}{\frac {\sin(2x)}{x^{2}}}}&=&\displaystyle {\lim _{x\rightarrow 0^{+}}{\frac {2\cos(2x)}{2x}}}\\&&\\&=&\displaystyle {\lim _{x\rightarrow 0^{+}}{\frac {\cos(2x)}{x}}.}\\\end{array}}}$

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

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

Final Answer:
(a)    ${\displaystyle 9}$
(b)    ${\displaystyle \infty }$
(c)    ${\displaystyle -{\frac {3}{2}}}$