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. 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 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} or 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.}

Step 2:
To figure out which one, we factor the denominator to get
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}=\lim_{x\rightarrow -3^+} \frac{x}{(x-3)(x+3)}.}
We are taking a right hand limit. So, we are looking at values of 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 x}
a little bigger than 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 -3.} (You can imagine values like 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 x=-2.9.} )
For these values, the numerator will be negative.
Also, for these values, 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 x-3} will be negative and 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 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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