009A Sample Final 1, Problem 1

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

(a) $\lim _{x\rightarrow -3}{\frac {x^{3}-9x}{6+2x}}$

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

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

Foundations:
Recall:
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 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{f'(x)}{g'(x)}}   is finite or  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 \pm \infty ,}
then 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{f(x)}{g(x)}\,=\,\lim_{x\rightarrow \infty} \frac{f'(x)}{g'(x)}.}

Solution:

(a)

Step 1:
We begin by factoring the numerator. 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 -3} \frac{x^3-9x}{6+2x}\,=\,\lim_{x\rightarrow -3}\frac{x(x-3)(x+3)}{2(x+3)}.}
So, we can cancel 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+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 just plug in Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle x=-3}   to get
$\lim _{x\rightarrow -3}{\frac {x^{3}-9x}{6+2x}}\,=\,{\frac {(-3)(-3-3)}{2}}\,=\,{\frac {18}{2}}\,=\,9.$

(b)

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

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 $x$ goes to negative infinity, we have ${\sqrt {x^{2}}}=-x.$
So, we have
Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\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 (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle \lim _{x\rightarrow -\infty }{\frac {3x}{\sqrt {4x^{2}+x+5}}}\,=\,\lim _{x\rightarrow -\infty }{\frac {-3}{\sqrt {4+{\frac {1}{x}}+{\frac {5}{x^{2}}}}}}.}
So, we have
$\lim _{x\rightarrow -\infty }{\frac {3x}{\sqrt {4x^{2}+x+5}}}={\frac {-3}{\sqrt {4}}}={\frac {-3}{2}}.$

Final Answer:
(a)  $9$
(b)  $+\infty$
(c)  Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle -{\frac {3}{2}}}

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