007A Sample Final 3, Problem 1 Detailed Solution

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

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

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
Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\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 {\lim _{x\rightarrow 0}{\bigg (}{\frac {\sin(5x)}{x}}{\bigg )}\lim _{x\rightarrow 0}(1+{\sqrt {1-x}})}\\&&\\&=&\displaystyle {5\lim _{x\rightarrow 0}{\bigg (}{\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
${\begin{array}{rcl}\displaystyle {-2}&=&\displaystyle {\lim _{x\rightarrow 8}{\bigg [}{\frac {xf(x)}{3}}{\bigg ]}}\\&&\\&=&\displaystyle {\frac {\displaystyle {\lim _{x\rightarrow 8}xf(x)}}{\displaystyle {\lim _{x\rightarrow 8}3}}}\\&&\\&=&\displaystyle {{\frac {\displaystyle {\lim _{x\rightarrow 8}xf(x)}}{3}}.}\end{array}}$

Step 2:
If we multiply both sides of the last equation by  $3,$  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  Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \lim_{x\rightarrow 8} f(x)}   in the last equation,
we get
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \lim_{x\rightarrow 8} f(x)=-\frac{3}{4}.}

(c)

Step 1:
First, we write
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\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}\cdot \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
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \begin{array}{rcl} \displaystyle{\lim_{x\rightarrow -\infty} \frac{\sqrt{9x^6-x}}{3x^3+4x}} & = & \displaystyle{\frac{\displaystyle{\lim_{x\rightarrow -\infty}} \sqrt{9-\frac{1}{x^5}}}{\displaystyle{\lim_{x\rightarrow -\infty}}\bigg(3+\frac{4}{x^2}\bigg)}}\\ &&\\ & = & \displaystyle{\frac{\sqrt{9}}{3}}\\ &&\\ & = & \displaystyle{1.} \end{array}}

Final Answer:
(a)    Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 10}
(b)    Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle -\frac{3}{4}}
(c)    Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 1}

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