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P: 2,004
 Quote by Nerpilis no I did not realize that $$\left(1+\frac{1}{n}\right)^n \left(1+ \frac{1}{n}\right)$$ 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 $$\lim_{n\rightarrow\infty}(\frac{n+3}{n+1})^{n+4} =$$ $$\lim_{n\rightarrow\infty}((\frac{n + 3}{n + 1})(\frac{1/n}{1/n}))^{n+4} =$$ $$\lim_{n\rightarrow\infty}\frac{(1 + (3/n))^{n+4}}{(1 + (1/n))^{n+4}} =$$ 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:
$$\lim_{n\to \infty} \left(1+\frac{1}{n}\right)^{n+4}=e^4$$
? 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: $\lim_{n\to \infty} (1+x/n)^n$ which looks alot like the one for e, what method would be useful? (Hint: substitution)

Another way is to write:
$$\frac{n+3}{n+1}=\frac{n+1+2}{n+1}=1+\frac{2}{n+1}$$