009C Sample Midterm 2, Problem 1

Evaluate:

(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 _{n\rightarrow \infty} \frac{1}{\big(\frac{n-4}{n}\big)^n}}

(b) $\sum _{n=1}^{\infty }{\frac {1}{2}}{\bigg (}{\frac {1}{4}}{\bigg )}^{n-1}$

Foundations:
1. L'Hôpital's Rule
Suppose that Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle \lim _{x\rightarrow \infty }f(x)}   and $\lim _{x\rightarrow \infty }g(x)$  are both zero or both $\pm \infty .$
If $\lim _{x\rightarrow \infty }{\frac {f'(x)}{g'(x)}}$  is finite or  $\pm \infty ,$
then $\lim _{x\rightarrow \infty }{\frac {f(x)}{g(x)}}\,=\,\lim _{x\rightarrow \infty }{\frac {f'(x)}{g'(x)}}.$
2. The sum of a convergent geometric series is   ${\frac {a}{1-r}}$
where $r$ is the ratio of the geometric series
and $a$ is the first term of the series.

Solution:

(a)

Step 1:
Let         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{y} & = & \displaystyle{\lim_{n\rightarrow \infty} \bigg(\frac{n-4}{n}\bigg)^n}\\ &&\\ & = & \displaystyle{\lim_{n\rightarrow \infty} \bigg(1-\frac{4}{n}\bigg)^n.} \end{array}}
We then take the natural log of both sides 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 \ln y = \ln\bigg(\lim_{n\rightarrow \infty} \bigg(1-\frac{4}{n}\bigg)^n\bigg).}

Step 2:
We can interchange limits and continuous functions.
Therefore, 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{\ln y} & = & \displaystyle{\lim_{n\rightarrow \infty} \ln \bigg(1-\frac{4}{n}\bigg)^n}\\ &&\\ & = & \displaystyle{\lim_{n\rightarrow \infty} n\ln\bigg(1-\frac{4}{n}\bigg)}\\ &&\\ & = & \displaystyle{\lim_{n\rightarrow \infty} \frac{\ln \bigg(1-\frac{4}{n}\bigg)}{\frac{1}{n}}.} \end{array}}
Now, this limit has the form 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{0}{0}.}
Hence, we can use L'Hopital's Rule to calculate this limit.

Step 3:
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{\ln y } & = & \displaystyle{\lim_{n\rightarrow \infty} \frac{\ln \bigg(1-\frac{4}{n}\bigg)}{\frac{1}{n}}}\\ &&\\ & = & \displaystyle{\lim_{x\rightarrow \infty} \frac{\ln \bigg(1-\frac{4}{x}\bigg)}{\frac{1}{x}}}\\ &&\\ & \overset{L'H}{=} & \displaystyle{\lim_{x\rightarrow \infty} \frac{\frac{1}{\big(1-\frac{4}{x}\big)}\frac{4}{x^2}}{\big(-\frac{1}{x^2}\big)}}\\ &&\\ & = & \displaystyle{\lim_{x\rightarrow \infty} \frac{-4x}{x-4}}\\ &&\\ & = & \displaystyle{-4.} \end{array}}

Step 4:
Since 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 \ln y= -4,} we know
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 y=e^{-4}.}
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 _{n\rightarrow \infty} \frac{1}{\big(\frac{n-4}{n}\big)^n}} & = & \displaystyle{\frac{\lim_{n\rightarrow \infty} 1}{\lim_{n\rightarrow \infty} \big(\frac{n-4}{n}\big)^n}}\\ &&\\ & = & \displaystyle{\frac{1}{e^{-4}}}\\ &&\\ & = & \displaystyle{e^4.} \end{array}}

(b)

Step 1:
First, we not that this is a geometric series with 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 r=\frac{1}{4}.}
Since 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 |r|=\frac{1}{4}<1,}
this series converges.

Step 2:
Now, we need to find the sum of this series.
The first term of the series is 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 a_1=\frac{1}{2}.}
Hence, the sum of the series is
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{\frac{a_1}{1-r}} & = & \displaystyle{\frac{\frac{1}{2}}{1-\frac{1}{4}}}\\ &&\\ & = & \displaystyle{\frac{\big(\frac{1}{2}\big)}{\big(\frac{3}{4}\big)}}\\ &&\\ & = & \displaystyle{\frac{2}{3}.} \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 e^{4}}
(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{2}{3}}

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