007A Sample Midterm 1, Problem 1 Detailed Solution

(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)

Find the following limits:

(a) Find  $\lim _{x\rightarrow 2}g(x),$ provided that  $\lim _{x\rightarrow 2}{\bigg [}{\frac {4-g(x)}{x}}{\bigg ]}=5.$ (b) Find  $\lim _{x\rightarrow 0}{\frac {\sin(4x)}{5x}}$ (c) Evaluate  $\lim _{x\rightarrow -3^{+}}{\frac {x}{x^{2}-9}}$ Background Information:
1. If  $\lim _{x\rightarrow a}g(x)\neq 0,$ we have
$\lim _{x\rightarrow a}{\frac {f(x)}{g(x)}}={\frac {\lim _{x\rightarrow a}f(x)}}{\lim _{x\rightarrow a}g(x)}}}.$ 2. Recall
$\lim _{x\rightarrow 0}{\frac {\sin x}{x}}=1$ Solution:

(a)

Step 1:
Since  $\lim _{x\rightarrow 2}x=2\neq 0,$ we have
${\begin{array}{rcl}\displaystyle {5}&=&\displaystyle {\lim _{x\rightarrow 2}{\bigg [}{\frac {4-g(x)}{x}}{\bigg ]}}\\&&\\&=&\displaystyle {\frac {\lim _{x\rightarrow 2}(4-g(x))}}{\lim _{x\rightarrow 2}x}}}\\&&\\&=&\displaystyle {{\frac {\lim _{x\rightarrow 2}(4-g(x))}}{2}}.}\end{array}}$ Step 2:
If we multiply both sides of the last equation by  $2,$ we get
$10=\lim _{x\rightarrow 2}(4-g(x)).$ Now, using linearity properties of limits, we have
${\begin{array}{rcl}\displaystyle {10}&=&\displaystyle {\lim _{x\rightarrow 2}4-\lim _{x\rightarrow 2}g(x)}\\&&\\&=&\displaystyle {4-\lim _{x\rightarrow 2}g(x).}\\\end{array}}$ Step 3:
Solving for  $\lim _{x\rightarrow 2}g(x)$ in the last equation,
we get

$\lim _{x\rightarrow 2}g(x)=-6.$ (b)

Step 1:
First, we write
$\lim _{x\rightarrow 0}{\frac {\sin(4x)}{5x}}=\lim _{x\rightarrow 0}{\bigg (}{\frac {4}{5}}\cdot {\frac {\sin(4x)}{4x}}{\bigg )}.$ Step 2:
Now, we have
${\begin{array}{rcl}\displaystyle {\lim _{x\rightarrow 0}{\frac {\sin(4x)}{5x}}}&=&\displaystyle {{\frac {4}{5}}\lim _{x\rightarrow 0}{\frac {\sin(4x)}{4x}}}\\&&\\&=&\displaystyle {{\frac {4}{5}}(1)}\\&&\\&=&\displaystyle {{\frac {4}{5}}.}\end{array}}$ (c)

Step 1:
When we plug in values close to  $-3$ into   ${\frac {x}{x^{2}-9}},$ we get a small denominator, which results in a large number.
Thus,
$\lim _{x\rightarrow -3^{+}}{\frac {x}{x^{2}-9}}$ is either equal to  $\infty$ or  $-\infty .$ Step 2:
To figure out which one, we factor the denominator to get
$\lim _{x\rightarrow -3^{+}}{\frac {x}{x^{2}-9}}=\lim _{x\rightarrow -3^{+}}{\frac {x}{(x-3)(x+3)}}.$ We are taking a right hand limit. So, we are looking at values of  $x$ a little bigger than  $-3.$ (You can imagine values like  $x=-2.9.$ )
For these values, the numerator will be negative.
Also, for these values,  $x-3$ will be negative and  $x+3$ will be positive.
Therefore, the denominator will be negative.
Since both the numerator and denominator will be negative (have the same sign),
$\lim _{x\rightarrow -3^{+}}{\frac {x}{x^{2}-9}}=\infty .$ (a)     $-6$ (b)     ${\frac {4}{5}}$ (c)     $\infty$ 