009A Sample Final 1, Problem 1 Detailed Solution

In each part, compute the limit. If the limit is infinite, be sure to specify positive or negative infinity.

(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 \lim_{x\rightarrow -3} \frac{x^3-9x}{6+2x}}

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

(c)   $\lim _{x\rightarrow -\infty }{\frac {3x}{\sqrt {4x^{2}+x+5}}}$

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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  Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle 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 factoring the numerator. We have
$\lim _{x\rightarrow -3}{\frac {x^{3}-9x}{6+2x}}\,=\,\lim _{x\rightarrow -3}{\frac {x(x-3)(x+3)}{2(x+3)}}.$
So, we can cancel  $x+3$  in the numerator and denominator. Thus, we have
$\lim _{x\rightarrow -3}{\frac {x^{3}-9x}{6+2x}}\,=\,\lim _{x\rightarrow -3}{\frac {x(x-3)}{2}}.$

Step 2:
Now, we can plug in  $x=-3$  to get
${\begin{array}{rcl}\displaystyle {\lim _{x\rightarrow -3}{\frac {x^{3}-9x}{6+2x}}}&=&\displaystyle {\frac {(-3)(-3-3)}{2}}\\&&\\&=&\displaystyle {\frac {18}{2}}\\&&\\&=&\displaystyle {9.}\end{array}}$

(b)

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

Step 2:
This limit is   $\infty .$

(c)

Step 1:
We have
$\lim _{x\rightarrow -\infty }{\frac {3x}{\sqrt {4x^{2}+x+5}}}=\lim _{x\rightarrow -\infty }{\frac {3x}{{\sqrt {x^{2}(4+{\frac {1}{x}}+{\frac {5}{x^{2}}}}})}}.$
Since we are looking at the limit as  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}   goes to negative infinity, 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 \sqrt{x^2}=-x.}
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 \lim_{x\rightarrow -\infty} \frac{3x}{\sqrt{4x^2+x+5}}\,=\,\lim_{x\rightarrow -\infty} \frac{3x}{-x\sqrt{4+\frac{1}{x}+\frac{5}{x^2}}}.}

Step 2:
We simplify 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 -\infty} \frac{3x}{\sqrt{4x^2+x+5}}} & = & \displaystyle{\lim_{x\rightarrow -\infty} \frac{-3}{\sqrt{4+\frac{1}{x}+\frac{5}{x^2}}}}\\ &&\\ & = & \displaystyle{-\frac{3}{\sqrt{4}}}\\ &&\\ & = & \displaystyle{-\frac{3}{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 9}
(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 \infty}
(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{3}{2}}

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