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

(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  $g$  are differentiable functions
on an open interval  $I$  containing  $c,$  and  $g'(x)\neq 0$  on  $I$  except possibly at  $c.$
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 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 just 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
${\begin{array}{rcl}\displaystyle {\lim _{x\rightarrow 0^{+}}{\frac {\sin(2x)}{x^{2}}}}&=&\displaystyle {\lim _{x\rightarrow 0^{+}}{\frac {2\cos(2x)}{2x}}}\\&&\\&=&\displaystyle {\lim _{x\rightarrow 0^{+}}{\frac {\cos(2x)}{x}}.}\\\end{array}}$

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
$\lim _{x\rightarrow -\infty }{\frac {3x}{\sqrt {4x^{2}+x+5}}}\,=\,\lim _{x\rightarrow -\infty }{\frac {3x}{-x{\sqrt {4+{\frac {1}{x}}+{\frac {5}{x^{2}}}}}}}.$

Step 2:
We simplify to get
${\begin{array}{rcl}\displaystyle {\lim _{x\rightarrow -\infty }{\frac {3x}{\sqrt {4x^{2}+x+5}}}}&=&\displaystyle {\lim _{x\rightarrow -\infty }{\frac {-3}{\sqrt {4+{\frac {1}{x}}+{\frac {5}{x^{2}}}}}}}\\&&\\&=&\displaystyle {-{\frac {3}{\sqrt {4}}}}\\&&\\&=&\displaystyle {-{\frac {3}{2}}.}\end{array}}$

Final Answer:
(a)    $9$
(b)    $\infty$
(c)    $-{\frac {3}{2}}$

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