009A Sample Final 2, Problem 1

Compute

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

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

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

Foundations:
L'Hôpital's Rule
Suppose that  $\lim _{x\rightarrow \infty }f(x)$  and  $\lim _{x\rightarrow \infty }g(x)$  are both zero or both  $\pm \infty .$
If  $\lim _{x\rightarrow \infty }{\frac {f'(x)}{g'(x)}}$  is finite or  $\pm \infty ,$
then  $\lim _{x\rightarrow \infty }{\frac {f(x)}{g(x)}}\,=\,\lim _{x\rightarrow \infty }{\frac {f'(x)}{g'(x)}}.$

Solution:

(a)

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

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

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

Final Answer:
(a)   ${\frac {1}{6}}$
(b)   $0$
(c)   $-{\frac {1}{2}}$

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