009A Sample Final 2, Problem 1 Detailed Solution

Compute

(a)  $\lim _{x\rightarrow 4}{\frac {{\sqrt {x+5}}-3}{x-4}}$

(b)  $\lim _{x\rightarrow 0}{\frac {\sin ^{2}x}{3x}}$

(c)  $\lim _{x\rightarrow -\infty }{\frac {\sqrt {x^{2}+2}}{2x-1}}$

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Background Information:
L'Hôpital's Rule, Part 1
Let  Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle \lim _{x\rightarrow c}f(x)=0}   and  Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle \lim _{x\rightarrow c}g(x)=0,}   where  $f$  and  $g$  are differentiable functions
on an open interval  $I$  containing  Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle c,}   and  $g'(x)\neq 0$  on  $I$  except possibly at  $c.$
Then,   Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\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 noticing that if we plug in  $x=4$  into
${\frac {{\sqrt {x+5}}-3}{x-4}},$
we get   ${\frac {0}{0}}.$

Step 2:
Now, we multiply the numerator and denominator by the conjugate of the numerator.
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 4}{\frac {{\sqrt {x+5}}-3}{x-4}}}&=&\displaystyle {\lim _{x\rightarrow 4}{\frac {{\sqrt {x+5}}-3}{x-4}}\cdot {\frac {({\sqrt {x+5}}+3)}{({\sqrt {x+5}}+3)}}}\\&&\\&=&\displaystyle {\lim _{x\rightarrow 4}{\frac {(x+5)-9}{(x-4)({\sqrt {x+5}}+3)}}}\\&&\\&=&\displaystyle {\lim _{x\rightarrow 4}{\frac {x-4}{(x-4)({\sqrt {x+5}}+3)}}}\\&&\\&=&\displaystyle {\lim _{x\rightarrow 4}{\frac {1}{{\sqrt {x+5}}+3}}}\\&&\\&=&\displaystyle {\frac {1}{{\sqrt {9}}+3}}\\&&\\&=&\displaystyle {{\frac {1}{6}}.}\end{array}}}

(b)

Step 1:
We proceed using L'Hôpital's Rule. So, 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 0} \frac{\sin^2 (x)}{3x}} & \overset{L'H}{=} & \displaystyle{\lim_{x\rightarrow 0}\frac{2\sin(x)\cos(x)}{3}.} \end{array}}

Step 2:
Now, we plug in  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 x=0}   to 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 \begin{array}{rcl} \displaystyle{\lim_{x\rightarrow 0} \frac{\sin^2 (x)}{3x}} & = & \displaystyle{\frac{2\sin(0)\cos(0)}{3}}\\ &&\\ & = & \displaystyle{\frac{2(0)(1)}{3}}\\ &&\\ & = & \displaystyle{0.} \end{array}}

(c)

Step 1:
First, 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{x^2+2}}{2x-1}} & = & \displaystyle{\lim_{x\rightarrow -\infty} \frac{\sqrt{x^2(1+\frac{2}{x^2})}}{2x-1}}\\ &&\\ & = & \displaystyle{\lim_{x\rightarrow -\infty} \frac{|x|\sqrt{1+\frac{2}{x^2}}}{2x-1}}\\ &&\\ & = & \displaystyle{\lim_{x\rightarrow -\infty} \frac{-x\sqrt{1+\frac{2}{x^2}}}{x(2-\frac{1}{x})}}\\ &&\\ & = & \displaystyle{\lim_{x\rightarrow -\infty} \frac{-1\sqrt{1+\frac{2}{x^2}}}{(2-\frac{1}{x})}.} \end{array}}

Step 2:
Now,
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{x^2+2}}{2x-1}} & = & \displaystyle{\lim_{x\rightarrow -\infty} \frac{-1\sqrt{1+\frac{2}{x^2}}}{2-\frac{1}{x}} }\\ &&\\ & = & \displaystyle{\frac{-\sqrt{1+0}}{2-0}}\\ &&\\ & = & \displaystyle{-\frac{1}{2}.} \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 \frac{1}{6}}
(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 0}
(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 -\frac{1}{2}}

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