Quote by Nerpilis
no I did not realize that [tex]\left(1+\frac{1}{n}\right)^n \left(1+ \frac{1}{n}\right)[/tex] I guess that would make the limit equal to e^2?

No. Why don't you just write it out and use some limit rules. This looks like the limit of a product of 2 factors, both with limits you know.
as far as the second problem I don't know what exp(x) is but i recognized that I can get things to look more like e
[tex]\lim_{n\rightarrow\infty}(\frac{n+3}{n+1})^{n+4} = [/tex]
[tex]\lim_{n\rightarrow\infty}((\frac{n + 3}{n + 1})(\frac{1/n}{1/n}))^{n+4} = [/tex]
[tex]\lim_{n\rightarrow\infty}\frac{(1 + (3/n))^{n+4}}{(1 + (1/n))^{n+4}} =[/tex]
from here the denominator looks similar to e and the numerator almost looks like e, however I’m a little hesitant to call the limit e/(e^4) = 1/(e^3)

Don't just guess what you think the limit should be. What makes you think that:
[tex]\lim_{n\to \infty} \left(1+\frac{1}{n}\right)^{n+4}=e^4[/tex]
? You can solve this limit in the same way as the previous one by considering it a product of 2 limits.
For a limit like: [itex]\lim_{n\to \infty} (1+x/n)^n[/itex] which looks alot like the one for e, what method would be useful? (Hint: substitution)
Another way is to write:
[tex]\frac{n+3}{n+1}=\frac{n+1+2}{n+1}=1+\frac{2}{n+1}[/tex]