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 $\lim _{x\rightarrow -3^{+}}{\frac {x}{x^{2}-9}}$

Foundations:
1. Linearity rules of limits
2. Limit sin(x)/x
3. Left and right hand limits

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),
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}=+\infty.}

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

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