find the [tex]\lim_{n\rightarrow\infty}(\frac{n+1}{n})^{n+1} = [/tex](adsbygoogle = window.adsbygoogle || []).push({});

so far what i have is

[tex]\lim_{n\rightarrow\infty}(\frac{n+1}{n})^{n+1} = \lim_{n\rightarrow\infty}(\frac{n}{n} + \frac{1}{n})^{n+1} = \lim_{n\rightarrow\infty}(1 + \frac{1}{n})^{n+1}\\[/tex]

I know this has got to go to [tex]e[/tex] or something very close, my question lies in if I should be concerned with the +1 in the n+1 in the exponet.

There is another one which I'm totaly stumped on for a starting point, my gut tells me the limit is 1 or it may not even exist altogether.

[tex]\lim_{n\rightarrow\infty}(\frac{n+3}{n+1})^{n+4} = [/tex]

but then again it could be +[tex]\infty[/tex] if you realize that in the () the numerator is larger than the denominator such that as the exponet is applied it goes to infinity instead od zero.

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# Homework Help: Limit questions

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