009C Sample Midterm 3, Problem 1 Detailed Solution

Test if the following sequence ${a_{n}}$ converges or diverges.

If it converges, also find the limit of the sequence.

$a_{n}=\left({\frac {n-7}{n}}\right)^{\frac {1}{n}}$

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Background Information:
L'Hôpital's Rule, Part 2
Let  $f$  and  $g$  be differentiable functions on the open interval  $(a,\infty )$  for some value  $a,$
where  $g'(x)\neq 0$  on  $(a,\infty )$  and  $\lim _{x\rightarrow \infty }{\frac {f(x)}{g(x)}}$  returns either  ${\frac {0}{0}}$  or  ${\frac {\infty }{\infty }}.$
Then,   $\lim _{x\rightarrow \infty }{\frac {f(x)}{g(x)}}=\lim _{x\rightarrow \infty }{\frac {f'(x)}{g'(x)}}.$

Solution:

Step 1:
First, notice that
$\lim _{n\rightarrow \infty }\ln n=\infty$
and
$\lim _{n\rightarrow \infty }n=\infty .$
Therefore, the limit has the form  ${\frac {\infty }{\infty }},$
which means that we can use L'Hopital's Rule to calculate this limit.

Step 2:
First, switch to the variable  $x$   so that we have functions and
can take derivatives. Thus, using L'Hopital's Rule, we have
${\begin{array}{rcl}\displaystyle {\lim _{n\rightarrow \infty }{\frac {\ln n}{n}}}&=&\displaystyle {\lim _{x\rightarrow \infty }{\frac {\ln x}{x}}}\\&&\\&{\overset {L'H}{=}}&\displaystyle {\lim _{x\rightarrow \infty }{\frac {{\big (}{\frac {1}{x}}{\big )}}{1}}}\\&&\\&=&\displaystyle {0.}\end{array}}$

Final Answer:
The sequence converges. The limit of the sequence is  $0.$

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