009A Sample Midterm 2, Problem 1

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

Foundations:
1. lim sinx/x
2. Left and right hand limit

Solution:

(a)

Step 1:
We begin by noticing that 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:

Step 2:

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

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