009A Sample Midterm 2, Problem 1 Detailed Solution

Evaluate the following limits.

(a) Find  $\lim _{x\rightarrow 2}{\frac {{\sqrt {x^{2}+12}}-4}{x-2}}$

(b) Find  $\lim _{x\rightarrow 0}{\frac {\sin(3x)}{\sin(7x)}}$

(c) Evaluate  $\lim _{x\rightarrow ({\frac {\pi }{2}})^{-}}\tan(x)$

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Background Information:
Recall
$\lim _{x\rightarrow 0}{\frac {\sin x}{x}}=1$

Solution:

(a)

Step 1:
We begin by noticing that if we plug in  $x=2$  into
${\frac {{\sqrt {x^{2}+12}}-4}{x-2}},$
we get   ${\frac {0}{0}}.$

Step 2:
Now, we multiply the numerator and denominator by the conjugate of the numerator.
Hence, we have
${\begin{array}{rcl}\displaystyle {\lim _{x\rightarrow 2}{\frac {{\sqrt {x^{2}+12}}-4}{x-2}}}&=&\displaystyle {\lim _{x\rightarrow 2}{\frac {({\sqrt {x^{2}+12}}-4)}{(x-2)}}{\frac {({\sqrt {x^{2}+12}}+4)}{({\sqrt {x^{2}+12}}+4)}}}\\&&\\&=&\displaystyle {\lim _{x\rightarrow 2}{\frac {(x^{2}+12)-16}{(x-2)({\sqrt {x^{2}+12}}+4)}}}\\&&\\&=&\displaystyle {\lim _{x\rightarrow 2}{\frac {x^{2}-4}{(x-2)({\sqrt {x^{2}+12}}+4)}}}\\&&\\&=&\displaystyle {\lim _{x\rightarrow 2}{\frac {(x-2)(x+2)}{(x-2)({\sqrt {x^{2}+12}}+4)}}}\\&&\\&=&\displaystyle {\lim _{x\rightarrow 2}{\frac {x+2}{{\sqrt {x^{2}+12}}+4}}}\\&&\\&=&\displaystyle {\frac {4}{8}}\\&&\\&=&\displaystyle {{\frac {1}{2}}.}\end{array}}$

(b)

Step 1:
First, we write
${\begin{array}{rcl}\displaystyle {\lim _{x\rightarrow 0}{\frac {\sin(3x)}{\sin(7x)}}}&=&\displaystyle {\lim _{x\rightarrow 0}{\frac {\sin(3x)}{x}}{\frac {x}{\sin(7x)}}}\\&&\\&=&\displaystyle {\lim _{x\rightarrow 0}{\frac {3}{7}}{\frac {\sin(3x)}{3x}}{\frac {7x}{\sin(7x)}}}\\&&\\&=&\displaystyle {{\frac {3}{7}}\lim _{x\rightarrow 0}{\frac {\sin(3x)}{3x}}{\frac {7x}{\sin(7x)}}.}\end{array}}$

Step 2:
Now, we have
${\begin{array}{rcl}\displaystyle {\lim _{x\rightarrow 0}{\frac {\sin(3x)}{\sin(7x)}}}&=&\displaystyle {{\frac {3}{7}}\lim _{x\rightarrow 0}{\frac {\sin(3x)}{3x}}{\frac {7x}{\sin(7x)}}}\\&&\\&=&\displaystyle {{\frac {3}{7}}{\bigg (}\lim _{x\rightarrow 0}{\frac {\sin(3x)}{3x}}{\bigg )}{\bigg (}\lim _{x\rightarrow 0}{\frac {7x}{\sin(7x)}}{\bigg )}}\\&&\\&=&\displaystyle {{\frac {3}{7}}(1)(1)}\\&&\\&=&\displaystyle {{\frac {3}{7}}.}\end{array}}$

(c)

Step 1:
First, we write
$\lim _{x\rightarrow {\big (}{\frac {\pi }{2}}{\big )}^{-}}\tan x=\lim _{x\rightarrow {\big (}{\frac {\pi }{2}}{\big )}^{-}}{\frac {\sin x}{\cos x}}.$

Step 2:
When we plug in values a little smaller than  ${\frac {\pi }{2}}$  into   ${\frac {\sin x}{\cos x}},$
we get a small denominator, which results in a large number.
Thus,
$\lim _{x\rightarrow {\big (}{\frac {\pi }{2}}{\big )}^{-}}{\frac {\sin x}{\cos x}}$
is either equal to  $\infty$  or  $-\infty .$

Step 3:
Since we are calculating a left hand limit, we are considering angles in the first quadrant.
In the first quadrant,  $\sin x$  and  $\cos x$  are both positive.
Hence,
$\lim _{x\rightarrow {\big (}{\frac {\pi }{2}}{\big )}^{-}}{\frac {\sin x}{\cos x}}=\infty .$

Final Answer:
(a)     ${\frac {1}{2}}$
(b)     ${\frac {3}{7}}$
(c)     $\infty$

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