009A Sample Midterm 1, Problem 1

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 Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \lim _{x\rightarrow -3^+} \frac{x}{x^2-9} }

Foundations:
1. If $\lim _{x\rightarrow a}g(x)\neq 0$, we have
$\lim _{x\rightarrow a}{\frac {f(x)}{g(x)}}={\frac {\displaystyle {\lim _{x\rightarrow a}f(x)}}{\displaystyle {\lim _{x\rightarrow a}g(x)}}}.$
2. $\lim _{x\rightarrow 0}{\frac {\sin x}{x}}=1$

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
$\lim _{x\rightarrow 0}{\frac {\sin(4x)}{5x}}=\lim _{x\rightarrow 0}{\frac {4}{5}}{\bigg (}{\frac {\sin(4x)}{4x}}{\bigg )}.$

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 $-3$ into ${\frac {x}{x^{2}-9}},$
we get ${\frac {-3}{0}}.$
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),
$\lim _{x\rightarrow -3^{+}}{\frac {x}{x^{2}-9}}=+\infty .$

Final Answer:
(a)     $\lim _{x\rightarrow 2}g(x)=-6$
(b)     ${\frac {4}{5}}$
(c)     $+\infty$

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