A_n = (1 + (3/n) )^4n
I don't know hot to find if this converges. I feel like I should break up the exponent, but I don't know how that would help.
This uses a couple of neat tricks.
Let this limit equal A. Then take the natural log of both sides and you get:
$\displaystyle \ln(A)=4n\ln(1+\frac{3}{n})$
$\displaystyle \frac{\ln(1+\frac{3}{n})}{(4n)^{-1}}=\ln(A)$
Now if you try to evaluate this limit for n -> infinity, you get an indeterminate form, so you can use L'Hopitals rule.
$\displaystyle \frac{(\frac{n}{n+3}) (\frac{-3}{n^2})}{\frac{-4}{n^2}}=\ln(A)$
The right hand side stays the same because after we apply the rule, nothing changes about the limit.
Now it's just simplifying and you should get it down to:
$\displaystyle \frac{3n}{4n+12}=\ln(A)$
Raise both sides to the power of e and you get $\displaystyle A=e^{\frac{3}{4}}$
Hello, veronicak5678!
We're expected to know this fact:
. . $\displaystyle \lim_{z\to\infty}\left(1 + \tfrac{1}{z}\right)^z \;=\;e$
We want: .$\displaystyle \lim_{n\to\infty}\left(1 + \tfrac{3}{n}\right)^{4n} $$\displaystyle A_n \:= \:\left(1 + \tfrac{3}{n}\right)^{4n}$
Multiply the exponent by $\displaystyle \tfrac{3}{3}$
. . $\displaystyle \left(1 + \tfrac{3}{n}\right)^{4n\cdot\frac{3}{3}} \;=\;\left(1 + \tfrac{3}{n}\right)^{\frac{n}{3}\cdot12} \;=\; \bigg[(1 + \tfrac{3}{n})^{\frac{n}{3}}\bigg]^{12} \;= \bigg[\left(1 + \frac{1}{\frac{n}{3}}\right)^{\frac{n}{3}}\bigg]^{12} $
$\displaystyle \text{Let }\:z \,=\,\tfrac{n}{3}$
. . and we have: .$\displaystyle \lim_{z\to\infty}\bigg[\left(1 + \frac{1}{z}\right)\bigg]^{12} \;=\;\bigg[\lim_{z\to\infty}\left(1 + \tfrac{1}{z}\right)^z\bigg]^{12} \;=\;e^{12}$