# 009A Sample Midterm 2, Problem 1 Detailed Solution

Evaluate the following limits.

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

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

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

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

Solution:

(a)

Step 1:
We begin by noticing that if we plug in  ${\displaystyle x=2}$  into
${\displaystyle {\frac {{\sqrt {x^{2}+12}}-4}{x-2}},}$
we get   ${\displaystyle {\frac {0}{0}}.}$
Step 2:
Now, we multiply the numerator and denominator by the conjugate of the numerator.
Hence, we have
${\displaystyle {\begin{array}{rcl}\displaystyle {\lim _{x\rightarrow 2}{\frac {{\sqrt {x^{2}+12}}-4}{x-2}}}&=&\displaystyle {\lim _{x\rightarrow 2}{\frac {({\sqrt {x^{2}+12}}-4)}{(x-2)}}{\frac {({\sqrt {x^{2}+12}}+4)}{({\sqrt {x^{2}+12}}+4)}}}\\&&\\&=&\displaystyle {\lim _{x\rightarrow 2}{\frac {(x^{2}+12)-16}{(x-2)({\sqrt {x^{2}+12}}+4)}}}\\&&\\&=&\displaystyle {\lim _{x\rightarrow 2}{\frac {x^{2}-4}{(x-2)({\sqrt {x^{2}+12}}+4)}}}\\&&\\&=&\displaystyle {\lim _{x\rightarrow 2}{\frac {(x-2)(x+2)}{(x-2)({\sqrt {x^{2}+12}}+4)}}}\\&&\\&=&\displaystyle {\lim _{x\rightarrow 2}{\frac {x+2}{{\sqrt {x^{2}+12}}+4}}}\\&&\\&=&\displaystyle {\frac {4}{8}}\\&&\\&=&\displaystyle {{\frac {1}{2}}.}\end{array}}}$

(b)

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

${\displaystyle {\begin{array}{rcl}\displaystyle {\lim _{x\rightarrow 0}{\frac {\sin(3x)}{\sin(7x)}}}&=&\displaystyle {{\frac {3}{7}}\lim _{x\rightarrow 0}{\frac {\sin(3x)}{3x}}{\frac {7x}{\sin(7x)}}}\\&&\\&=&\displaystyle {{\frac {3}{7}}{\bigg (}\lim _{x\rightarrow 0}{\frac {\sin(3x)}{3x}}{\bigg )}{\bigg (}\lim _{x\rightarrow 0}{\frac {7x}{\sin(7x)}}{\bigg )}}\\&&\\&=&\displaystyle {{\frac {3}{7}}(1)(1)}\\&&\\&=&\displaystyle {{\frac {3}{7}}.}\end{array}}}$

(c)

Step 1:
First, we write
${\displaystyle \lim _{x\rightarrow {\big (}{\frac {\pi }{2}}{\big )}^{-}}\tan x=\lim _{x\rightarrow {\big (}{\frac {\pi }{2}}{\big )}^{-}}{\frac {\sin x}{\cos x}}.}$
Step 2:
When we plug in values a little smaller than  ${\displaystyle {\frac {\pi }{2}}}$  into   ${\displaystyle {\frac {\sin x}{\cos x}},}$
we get a small denominator, which results in a large number.
Thus,
${\displaystyle \lim _{x\rightarrow {\big (}{\frac {\pi }{2}}{\big )}^{-}}{\frac {\sin x}{\cos x}}}$
is either equal to  ${\displaystyle \infty }$  or  ${\displaystyle -\infty .}$
Step 3:
Since we are calculating a left hand limit, we are considering angles in the first quadrant.
In the first quadrant,  ${\displaystyle \sin x}$  and  ${\displaystyle \cos x}$  are both positive.
Hence,
${\displaystyle \lim _{x\rightarrow {\big (}{\frac {\pi }{2}}{\big )}^{-}}{\frac {\sin x}{\cos x}}=\infty .}$

(a)     ${\displaystyle {\frac {1}{2}}}$
(b)     ${\displaystyle {\frac {3}{7}}}$
(c)     ${\displaystyle \infty }$