009A Sample Final 2, Problem 8 Detailed Solution

Compute

(a)  $\lim _{x\rightarrow \infty }{\frac {x^{-1}+x}{1+{\sqrt {1+x}}}}$

(b)  $\lim _{x\rightarrow 0}{\frac {\sin x}{\cos x-1}}$

(c)  $\lim _{x\rightarrow 1}{\frac {x^{3}-1}{x^{10}-1}}$

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Background Information:
L'Hôpital's Rule, Part 1
Let  $\lim _{x\rightarrow c}f(x)=0$  and  $\lim _{x\rightarrow c}g(x)=0,$  where  $f$  and  $g$  are differentiable functions
on an open interval  $I$  containing  $c,$  and  $g'(x)\neq 0$  on  $I$  except possibly at  Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle c.}
Then,   $\lim _{x\rightarrow c}{\frac {f(x)}{g(x)}}=\lim _{x\rightarrow c}{\frac {f'(x)}{g'(x)}}.$

Solution:

(a)

Step 1:
First, 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 \infty }{\frac {x^{-1}+x}{1+{\sqrt {1+x}}}}}&=&\displaystyle {\lim _{x\rightarrow \infty }{\frac {{\frac {1}{x}}+x}{1+{\sqrt {1+x}}}}}\\&&\\&=&\displaystyle {\lim _{x\rightarrow \infty }{\frac {{\frac {1}{x}}+x}{1+{\sqrt {1+x}}}}\cdot {\frac {{\big (}{\frac {1}{\sqrt {x}}}{\big )}}{{\big (}{\frac {1}{\sqrt {x}}}{\big )}}}}\\&&\\&=&\displaystyle {\lim _{x\rightarrow \infty }{\frac {{\frac {1}{x^{3/2}}}+{\sqrt {x}}}{{\frac {1}{\sqrt {x}}}+{\sqrt {{\frac {1}{x}}+1}}}}.}\end{array}}}

Step 2:
Now, 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 \infty }{\frac {x^{-1}+x}{1+{\sqrt {1+x}}}}}&=&\displaystyle {\lim _{x\rightarrow \infty }{\frac {{\frac {1}{x^{3/2}}}+{\sqrt {x}}}{{\frac {1}{\sqrt {x}}}+{\sqrt {{\frac {1}{x}}+1}}}}}\\&&\\&=&\displaystyle {\frac {\displaystyle {\lim _{x\rightarrow \infty }}{\bigg (}{\frac {1}{x^{3/2}}}+{\sqrt {x}}{\bigg )}}{\displaystyle {\lim _{x\rightarrow \infty }}{\bigg (}{\frac {1}{\sqrt {x}}}+{\sqrt {{\frac {1}{x}}+1}}{\bigg )}}}\\&&\\&=&\displaystyle {\frac {\displaystyle {\lim _{x\rightarrow \infty }}{\frac {1}{x^{3/2}}}+\displaystyle {\lim _{x\rightarrow \infty }}{\sqrt {x}}}{\displaystyle {\lim _{x\rightarrow \infty }}{\frac {1}{\sqrt {x}}}+\displaystyle {\lim _{x\rightarrow \infty }}{\sqrt {{\frac {1}{x}}+1}}}}\\&&\\&=&\displaystyle {\frac {0+\displaystyle {\lim _{x\rightarrow \infty }}{\sqrt {x}}}{0+1}}\\&&\\&=&\displaystyle {\infty .}\end{array}}}

(b)

Step 1:
First, we write
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 x}{\cos x-1}}}&=&\displaystyle {\lim _{x\rightarrow 0}{\frac {\sin x}{\cos x-1}}{\frac {(\cos x+1)}{(\cos x+1)}}}\\&&\\&=&\displaystyle {\lim _{x\rightarrow 0}{\frac {\sin x(\cos x+1)}{\cos ^{2}x-1}}}\\&&\\&=&\displaystyle {\lim _{x\rightarrow 0}{\frac {\sin x(\cos x+1)}{-\sin ^{2}x}}}\\&&\\&=&\displaystyle {\lim _{x\rightarrow 0}{\frac {\cos x+1}{-\sin x}}.}\end{array}}}

Step 2:
Now, 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 x}{\cos x-1}}}&=&\displaystyle {\lim _{x\rightarrow 0^{+}}{\frac {\cos x+1}{-\sin x}}}\\&&\\&=&\displaystyle {-\infty }\end{array}}}
and
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 x}{\cos x-1}}}&=&\displaystyle {\lim _{x\rightarrow 0^{-}}{\frac {\cos x+1}{-\sin x}}}\\&&\\&=&\displaystyle {\infty .}\end{array}}}
Therefore,
Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle \lim _{x\rightarrow 0}{\frac {\sin x}{\cos x-1}}={\text{DNE}}.}

(c)

Step 1:
We proceed using L'Hôpital's Rule. So, we have
${\begin{array}{rcl}\displaystyle {\lim _{x\rightarrow 1}{\frac {x^{3}-1}{x^{10}-1}}}&{\overset {L'H}{=}}&\displaystyle {\lim _{x\rightarrow 1}{\frac {3x^{2}}{10x^{9}}}.}\end{array}}$

Step 2:
Now, we have
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 \begin{array}{rcl} \displaystyle{\lim_{x\rightarrow 1} \frac{x^3-1}{x^{10}-1}} & \overset{L'H}{=} & \displaystyle{\lim_{x\rightarrow 1}\frac{3x^2}{10x^9}}\\ &&\\ & = & \displaystyle{\frac{3(1)^2}{10(1)^9}}\\ &&\\ & = & \displaystyle{\frac{3}{10}.} \end{array}}

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 \infty}
(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 \text{DNE}}
(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}{10}}

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