limit of sequence

[missyc8]

lim n ->infinity ((n+6)/(n+1) )^n

= (1+ 6/n) / 1 + 1/n)^n n->infinity .. ineterm. form

so, apply LH rule.. lim n-> infinity lny =n ln (n+6)/(n+1) = lim n -> infinity ln ((n+6)/(n+1) )/ (n^-1) = 0/ infinity ..indeterm form...

i get stuck here...you have to use chain rule? if yes..i am not getting the right answer
can someone help?

 

[galactus]

You do not need L'Hopital. Long divide your problem and get $\displaystyle \lim_{n\to \infty}\left(1+\frac{5}{n+1}\right)^{n}$

Note that this is a famous 'e' limit. Note the silialrity with $\displaystyle e=\lim_{n\to \infty}\left(1+\frac{1}{n+1}\right)^{n}$.

See your limit now?.

 

[soroban]

Hello, missyc8!

Galactus is absolutely correct,
. . and some Olympic-level gymnastics is required . . .

$\displaystyle \lim_{n\to\infty} \left(\frac{n+6}{n+1}\right)^n$

$\displaystyle \text{We will use the definition: }\;\lim_{x\to\infty}\left(1 + \frac{1}{x}\right)^x \;=\;e$


$\displaystyle \text{Using long division: }\;\frac{n+6}{n+1} \;=\;1 + \frac{5}{n+1}$

. . $\displaystyle \text{Then: }\;\left(\frac{n+6}{n+1}\right)^n \;=\;\left(1 + \frac{5}{n+1}\right)^n$


$\displaystyle \text{Multiply top and bottom by }\left(1 + \tfrac{5}{n+1}\right)\!:\;\;\frac{\left(1 + \frac{5}{n+1}\right)^n}{1}\cdot\frac{1 + \frac{5}{n+1}}{1 + \frac{5}{n+1}} \;=\;\frac{\left(1 + \frac{5}{n+1}\right)^{n+1}}{1 + \frac{5}{n+1}}$

. . $\displaystyle \text{The numerator is: }\;\left(1 + \frac{5}{n+1}\right)^{n+1} \;=\;\;\left[\left(1 + \frac{5}{n+1}\right)^{n+1}\right]^{\frac{5}{5}} \;=\;\;\left[\left(1 + \frac{5}{n+1}\right)^{\frac{n+1}{5}}\right]^5 \;=\;\;\left[\left(1 + \frac{1}{\frac{n+1}{5}}\right)^{\frac{n+1}{5}}\right]^5$


$\displaystyle \text{And we have: }\;\lim_{n\to\infty}\frac{\left[\left(1 + \frac{1}{\frac{n+1}{5}}\right)^{\frac{n+1}{5}}\right]^5}{1 + \frac{5}{n+1}} \;\;=\;\;\frac{\lim\left[\left(1 + \frac{1}{\frac{n+1}{5}}\right)^{\frac{n+1}{5}}\right]^5} {\lim\left(1 + \frac{5}{n+1}\right)}$

. . . . . . . . $\displaystyle =\;\;\frac{\left[\lim\left(1 + \frac{1}{\frac{n+1}{5}}\right)^{\frac{n+1}{5}}\right]^5} {\lim\left(1 + \frac{5}{n+1}\right)} \;\;=\;\;\frac{e^5}{1+0} \;\;=\;\;e^5$

. . Whew! . . . I need a nap.

 

[Deleted member 4993]

However you could use the well known corollary - and avoid hurtful gymnastics:

$\displaystyle \lim_{n\to\infty}\left(1 + \frac{x}{n}\right)^n \;=\;e^x$

 

[missyc8]

thanks everyone for the help!

 

[BigGlenntheHeavy]

Another way.

$\displaystyle Find \ \lim_{n\to\infty}\bigg(\frac{n+6}{n+1}\bigg)^{n}, \ Let \ y \ = \ \lim_{n\to\infty}\bigg(\frac{n+6}{n+1}\bigg)^{n}$

$\displaystyle Then, \ ln|y| \ = \ ln\bigg[\lim_{n\to\infty}\bigg(\frac{n+6}{n+1}\bigg)^{n}\bigg].$

$\displaystyle ln|y| \ = \ \lim_{n\to\infty}ln\bigg(\frac{n+6}{n+1}\bigg)^{n} \ = \ \lim_{n\to\infty}nln\bigg(\frac{n+6}{n+1}\bigg)$

$\displaystyle \lim_{n\to\infty}nln\bigg(\frac{n+6}{n+1}\bigg) \ gives \ indeterminate \ form \ \infty \ * \ 0, \ however, \ not \ to \ despair.$

$\displaystyle \lim_{n\to\infty}\frac{ln\bigg(\frac{n+6}{n+1}\bigg)}{\frac{1}{n}} \ gives \ indeterminate \ form \ \frac{0}{0}.$

$\displaystyle Therefore, \ the \ Marqui \ to \ the \ rescue \ gives, \ \lim_{n\to\infty}nln\bigg(\frac{n+6}{n+1}\bigg) \ = \ 5.$

$\displaystyle Hence, \ ln|y| \ = \ 5, \ \implies \ y \ = \ e^{5}.$

$\displaystyle Therefore, \ after \ all \ is \ said \ and \ done, \ \lim_{n\to\infty}\bigg(\frac{n+6}{n+1}\bigg)^{n} \ = \ e^{5}.\ \ QED$