007A 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) $\lim _{x\rightarrow -\infty }{\frac {\sqrt {x^{2}+2}}{2x-1}}$

Background Information:
L'Hôpital's Rule, Part 1
Let $\lim _{x\rightarrow c}f(x)=0$ and $\lim _{x\rightarrow c}g(x)=0,$ where $f$ and $g$ are differentiable functions
on an open interval $I$ containing $c,$ and $g'(x)\neq 0$ on $I$ except possibly at $c.$
Then, $\lim _{x\rightarrow c}{\frac {f(x)}{g(x)}}=\lim _{x\rightarrow c}{\frac {f'(x)}{g'(x)}}.$

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
${\begin{array}{rcl}\displaystyle {\lim _{x\rightarrow 4}{\frac {{\sqrt {x+5}}-3}{x-4}}}&=&\displaystyle {\lim _{x\rightarrow 4}{\frac {{\sqrt {x+5}}-3}{x-4}}\cdot {\frac {({\sqrt {x+5}}+3)}{({\sqrt {x+5}}+3)}}}\\&&\\&=&\displaystyle {\lim _{x\rightarrow 4}{\frac {(x+5)-9}{(x-4)({\sqrt {x+5}}+3)}}}\\&&\\&=&\displaystyle {\lim _{x\rightarrow 4}{\frac {x-4}{(x-4)({\sqrt {x+5}}+3)}}}\\&&\\&=&\displaystyle {\lim _{x\rightarrow 4}{\frac {1}{{\sqrt {x+5}}+3}}}\\&&\\&=&\displaystyle {\frac {1}{{\sqrt {9}}+3}}\\&&\\&=&\displaystyle {{\frac {1}{6}}.}\end{array}}$

(b)

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

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

(c)

Step 1:

First, we have

{\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,
${\begin{array}{rcl}\displaystyle {\lim _{x\rightarrow -\infty }{\frac {\sqrt {x^{2}+2}}{2x-1}}}&=&\displaystyle {\lim _{x\rightarrow -\infty }{\frac {-1{\sqrt {1+{\frac {2}{x^{2}}}}}}{2-{\frac {1}{x}}}}}\\&&\\&=&\displaystyle {\frac {-{\sqrt {1+0}}}{2-0}}\\&&\\&=&\displaystyle {-{\frac {1}{2}}.}\end{array}}$

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