009A Sample Final 1, Problem 1

In each part, compute the limit. If the limit is infinite, be sure to specify positive or negative infinity.

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 -3} \frac{x^3-9x}{6+2x}}

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 \lim_{x\rightarrow 0^+} \frac{\sin (2x)}{x^2}}

c) $\lim _{x\rightarrow -\infty }{\frac {3x}{\sqrt {4x^{2}+x+5}}}$

Foundations:
Review L'Hopital's Rule

Solution:

(a)

Step 1:
We begin by factoring the numerator. We have
$\lim _{x\rightarrow -3}{\frac {x^{3}-9x}{6+2x}}=\lim _{x\rightarrow -3}{\frac {x(x-3)(x+3)}{2(x+3)}}$.
So, we can cancel $x+3$ in the numerator and denominator. Thus, we have
$\lim _{x\rightarrow -3}{\frac {x^{3}-9x}{6+2x}}=\lim _{x\rightarrow -3}{\frac {x(x-3)}{2}}$.

Step 2:
Now, we can just plug in $x=-3$ to get
$\lim _{x\rightarrow -3}{\frac {x^{3}-9x}{6+2x}}={\frac {(-3)(-3-3)}{2}}={\frac {18}{2}}=9$.

(b)

Step 1:
We proceed using L'Hopital's Rule. So, we have
Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle {\begin{array}{rcl}\displaystyle {\lim _{x\rightarrow 0^{+}}{\frac {\sin(2x)}{x^{2}}}}&=&\displaystyle {\lim _{x\rightarrow 0^{+}}{\frac {2\cos(2x)}{2x}}}\\&&\\&=&\displaystyle {\lim _{x\rightarrow 0^{+}}{\frac {\cos(2x)}{x}}}\\\end{array}}}

Step 2:
This limit is $+\infty$.

(c)

Step 1:
We have
Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle \lim _{x\rightarrow -\infty }{\frac {3x}{\sqrt {4x^{2}+x+5}}}=\lim _{x\rightarrow -\infty }{\frac {3x}{{\sqrt {x^{2}(4+{\frac {1}{x}}+{\frac {5}{x^{2}}}}})}}} .
Since we are looking at the limit as $x$ goes to negative infinity, we have ${\sqrt {x^{2}}}=-x$.
So, we have
Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle \lim _{x\rightarrow -\infty }{\frac {3x}{\sqrt {4x^{2}+x+5}}}=\lim _{x\rightarrow -\infty }{\frac {3x}{-x{\sqrt {4+{\frac {1}{x}}+{\frac {5}{x^{2}}}}}}}} .

Step 2:
We simplify to get
$\lim _{x\rightarrow -\infty }{\frac {3x}{\sqrt {4x^{2}+x+5}}}=\lim _{x\rightarrow -\infty }{\frac {-3}{\sqrt {4+{\frac {1}{x}}+{\frac {5}{x^{2}}}}}}$
So, we have
Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle \lim _{x\rightarrow -\infty }{\frac {3x}{\sqrt {4x^{2}+x+5}}}={\frac {-3}{\sqrt {4}}}={\frac {-3}{2}}} .

Final Answer:
(a) $9$.
(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 +\infty}
(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 \frac{-3}{2}}

Return to Sample Exam