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:
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 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
$\lim _{x\rightarrow -3}{\frac {x^{3}-9x}{6+2x}}\,=\,{\frac {(-3)(-3-3)}{2}}\,=\,{\frac {18}{2}}\,=\,9.$

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

Step 2:
We simplify to get
$\lim _{x\rightarrow -\infty }{\frac {3x}{\sqrt {4x^{2}+x+5}}}\,=\,\lim _{x\rightarrow -\infty }{\frac {-3}{\sqrt {4+{\frac {1}{x}}+{\frac {5}{x^{2}}}}}}.$
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)    $-{\frac {3}{2}}$

Return to Sample Exam