# Difference between revisions of "009A Sample Final 3, Problem 1"

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Find each of the following limits if it exists. If you think the limit does not exist provide a reason.

(a)  ${\displaystyle \lim _{x\rightarrow 0}{\frac {\sin(5x)}{1-{\sqrt {1-x}}}}}$

(b)  ${\displaystyle \lim _{x\rightarrow 8}f(x),}$  given that  ${\displaystyle \lim _{x\rightarrow 8}{\frac {xf(x)}{3}}=-2}$

(c)  ${\displaystyle \lim _{x\rightarrow -\infty }{\frac {\sqrt {9x^{6}-x}}{3x^{3}+4x}}}$

Foundations:
1. If  ${\displaystyle \lim _{x\rightarrow a}g(x)\neq 0,}$  we have
${\displaystyle \lim _{x\rightarrow a}{\frac {f(x)}{g(x)}}={\frac {\displaystyle {\lim _{x\rightarrow a}f(x)}}{\displaystyle {\lim _{x\rightarrow a}g(x)}}}.}$
2.  ${\displaystyle \lim _{x\rightarrow 0}{\frac {\sin x}{x}}=1}$

Solution:

(a)

Step 1:
We begin by noticing that we plug in  ${\displaystyle x=0}$  into
${\displaystyle {\frac {\sin(5x)}{1-{\sqrt {1-x}}}},}$
we get   ${\displaystyle {\frac {0}{0}}.}$
Step 2:
Now, we multiply the numerator and denominator by the conjugate of the denominator.
Hence, we have
${\displaystyle {\begin{array}{rcl}\displaystyle {\lim _{x\rightarrow 0}{\frac {\sin(5x)}{1-{\sqrt {1-x}}}}}&=&\displaystyle {\lim _{x\rightarrow 0}{\frac {\sin(5x)}{1-{\sqrt {1-x}}}}{\bigg (}{\frac {1+{\sqrt {1-x}}}{1+{\sqrt {1-x}}}}{\bigg )}}\\&&\\&=&\displaystyle {\lim _{x\rightarrow 0}{\frac {\sin(5x)(1+{\sqrt {1-x}})}{x}}}\\&&\\&=&\displaystyle {\lim _{x\rightarrow 0}{\frac {\sin(5x)}{x}}(1+{\sqrt {1-x}})}\\&&\\&=&\displaystyle {{\bigg (}\lim _{x\rightarrow 0}{\frac {\sin(5x)}{x}}{\bigg )}\lim _{x\rightarrow 0}(1+{\sqrt {1-x}})}\\&&\\&=&\displaystyle {{\bigg (}5\lim _{x\rightarrow 0}{\frac {\sin(5x)}{5x}}{\bigg )}(2)}\\&&\\&=&\displaystyle {5(1)(2)}\\&&\\&=&\displaystyle {10.}\end{array}}}$

(b)

Step 1:
Since  ${\displaystyle \lim _{x\rightarrow 8}3=3\neq 0,}$
we have
${\displaystyle {\begin{array}{rcl}\displaystyle {-2}&=&\displaystyle {\lim _{x\rightarrow 8}{\bigg [}{\frac {xf(x)}{3}}{\bigg ]}}\\&&\\&=&\displaystyle {\frac {\displaystyle {\lim _{x\rightarrow 8}xf(x)}}{\displaystyle {\lim _{x\rightarrow 8}3}}}\\&&\\&=&\displaystyle {{\frac {\displaystyle {\lim _{x\rightarrow 8}xf(x)}}{3}}.}\end{array}}}$
Step 2:
If we multiply both sides of the last equation by  ${\displaystyle 3,}$  we get
${\displaystyle -6=\lim _{x\rightarrow 8}xf(x).}$
Now, using properties of limits, we have
${\displaystyle {\begin{array}{rcl}\displaystyle {-6}&=&\displaystyle {{\bigg (}\lim _{x\rightarrow 8}x{\bigg )}{\bigg (}\lim _{x\rightarrow 8}f(x){\bigg )}}\\&&\\&=&\displaystyle {8\lim _{x\rightarrow 8}f(x).}\\\end{array}}}$
Step 3:
Solving for  ${\displaystyle \lim _{x\rightarrow 8}f(x)}$  in the last equation,
we get

${\displaystyle \lim _{x\rightarrow 8}f(x)=-{\frac {3}{4}}.}$

(c)

Step 1:
First, we write
${\displaystyle {\begin{array}{rcl}\displaystyle {\lim _{x\rightarrow -\infty }{\frac {\sqrt {9x^{6}-x}}{3x^{3}+4x}}}&=&\displaystyle {\lim _{x\rightarrow -\infty }{\frac {\sqrt {9x^{6}-x}}{3x^{3}+4x}}{\frac {{\big (}{\frac {1}{x^{3}}}{\big )}}{{\big (}{\frac {1}{x^{3}}}{\big )}}}}\\&&\\&=&\displaystyle {\lim _{x\rightarrow -\infty }{\frac {\sqrt {9-{\frac {1}{x^{5}}}}}{3+{\frac {4}{x^{2}}}}}.}\end{array}}}$
Step 2:
Now, we have
${\displaystyle {\begin{array}{rcl}\displaystyle {\lim _{x\rightarrow -\infty }{\frac {\sqrt {9x^{6}-x}}{3x^{3}+4x}}}&=&\displaystyle {\frac {\lim _{x\rightarrow -\infty }{\sqrt {9-{\frac {1}{x^{5}}}}}}{\lim _{x\rightarrow -\infty }3+{\frac {4}{x^{2}}}}}\\&&\\&=&\displaystyle {\frac {\sqrt {9}}{3}}\\&&\\&=&\displaystyle {1.}\end{array}}}$

Final Answer:
(a)    ${\displaystyle 10}$
(b)    ${\displaystyle -{\frac {3}{4}}}$
(c)    ${\displaystyle 1}$