# 007A 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  $\lim _{x\rightarrow 0}x^{2}\cos {\bigg (}{\frac {1}{x}}{\bigg )}$ Background Information:
1.  $\lim _{x\rightarrow 0}{\frac {\sin x}{x}}=1$ 2. Squeeze Theorem
Let  $f,g$ and  $h$ be functions on an open interval  $I$ containing  $c$ such that for all  $x$ in  $I,~f(x)\leq g(x)\leq h(x).$ If  $\lim _{x\rightarrow c}f(x)=L=\lim _{x\rightarrow c}h(x),$ then  $\lim _{x\rightarrow c}g(x)=L.$ 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
${\begin{array}{rcl}\displaystyle {\lim _{x\rightarrow 2}{\frac {{\sqrt {x^{2}+12}}-4}{x-2}}}&=&\displaystyle {\lim _{x\rightarrow 2}{\bigg [}{\frac {({\sqrt {x^{2}+12}}-4)}{(x-2)}}\cdot {\frac {({\sqrt {x^{2}+12}}+4)}{({\sqrt {x^{2}+12}}+4)}}{\bigg ]}}\\&&\\&=&\displaystyle {\lim _{x\rightarrow 2}{\frac {(x^{2}+12)-16}{(x-2)({\sqrt {x^{2}+12}}+4)}}}\\&&\\&=&\displaystyle {\lim _{x\rightarrow 2}{\frac {x^{2}-4}{(x-2)({\sqrt {x^{2}+12}}+4)}}}\\&&\\&=&\displaystyle {\lim _{x\rightarrow 2}{\frac {(x-2)(x+2)}{(x-2)({\sqrt {x^{2}+12}}+4)}}}\\&&\\&=&\displaystyle {\lim _{x\rightarrow 2}{\frac {x+2}{{\sqrt {x^{2}+12}}+4}}}\\&&\\&=&\displaystyle {\frac {4}{8}}\\&&\\&=&\displaystyle {{\frac {1}{2}}.}\end{array}}$ (b)

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

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

Step 1:
First, recall that
$-1\leq \cos(\theta )\leq 1$ for all  $\theta .$ Then, for all  $x\neq 0,$ $-1\leq \cos {\bigg (}{\frac {1}{x}}{\bigg )}\leq 1.$ Hence, for all  $x\neq 0,$ $-x^{2}\leq x^{2}\cos {\bigg (}{\frac {1}{x}}{\bigg )}\leq x^{2}.$ Step 2:
Now, notice
$\lim _{x\rightarrow 0}x^{2}=0$ and
$\lim _{x\rightarrow 0}-x^{2}=0.$ Step 3:
Since
$\lim _{x\rightarrow 0}x^{2}=\lim _{x\rightarrow 0}-x^{2}=0,$ we have
$\lim _{x\rightarrow 0}x^{2}\cos {\bigg (}{\frac {1}{x}}{\bigg )}=0$ by the Squeeze Theorem.

(a)     ${\frac {1}{2}}$ (b)     ${\frac {3}{7}}$ (c)     $0$ 