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)  $\lim _{x\rightarrow 0}{\frac {\sin(5x)}{1-{\sqrt {1-x}}}}$ (b)  $\lim _{x\rightarrow 8}f(x),$ given that  $\lim _{x\rightarrow 8}{\frac {xf(x)}{3}}=-2$ (c)  $\lim _{x\rightarrow -\infty }{\frac {\sqrt {9x^{6}-x}}{3x^{3}+4x}}$ Foundations:
1. If  $\lim _{x\rightarrow a}g(x)\neq 0,$ we have
$\lim _{x\rightarrow a}{\frac {f(x)}{g(x)}}={\frac {\lim _{x\rightarrow a}f(x)}}{\lim _{x\rightarrow a}g(x)}}}.$ 2.  $\lim _{x\rightarrow 0}{\frac {\sin x}{x}}=1$ Solution:

(a)

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

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

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

Step 1:
First, we write
${\begin{array}{rcl}\displaystyle {\lim _{x\rightarrow -\infty }{\frac {\sqrt {9x^{6}-x}}{3x^{3}+4x}}}&=&\displaystyle {\lim _{x\rightarrow -\infty }{\frac {\sqrt {9x^{6}(1-{\frac {1}{9x^{5}}})}}{3x^{3}+4x}}}\\&&\\&=&\displaystyle {\lim _{x\rightarrow -\infty }{\frac {3|x^{3}|{\sqrt {(1-{\frac {1}{9x^{5}}})}}}{3x^{3}+4x}}}\\&&\\&=&\displaystyle {\lim _{x\rightarrow -\infty }{\frac {3(-x^{3}){\sqrt {(1-{\frac {1}{9x^{5}}})}}}{3x^{3}(1+{\frac {4}{3x^{2}}})}}}\end{array}}$ Step 2:
Now, we have
${\begin{array}{rcl}\displaystyle {\lim _{x\rightarrow -\infty }{\frac {\sqrt {9x^{6}-x}}{3x^{3}+4x}}}&=&\displaystyle {\lim _{x\rightarrow -\infty }{\frac {3(-x^{3}){\sqrt {(1-{\frac {1}{9x^{5}}})}}}{3x^{3}(1+{\frac {4}{3x^{2}}})}}}\\&&\\&=&\displaystyle {\lim _{x\rightarrow -\infty }{\frac {-{\sqrt {(1-{\frac {1}{9x^{5}}})}}}{1+{\frac {4}{3x^{2}}}}}}\\&&\\&=&\displaystyle {\frac {-{\sqrt {1}}}{1}}\\&&\\&=&\displaystyle {-1.}\\\end{array}}$ (a)    $10$ (b)    $-{\frac {3}{4}}$ (c)    $-1$ 