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:
Review L'Hopital's Rule

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'Hopital'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) ${\frac {-3}{2}}$

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