# Find the limit of $2+\left(-\frac{2}{e}\right)^n$, as $n\to\infty$, if it exists

I'm absolutely unsure about how to approach this. I've considered changing it to $-2=\left(-\frac{2}{e}\right)^n$ and then using the properties of lograrithms, but $\ln(-2)$ is undefined, as is $\ln(-\frac{2}{e})$.

I can't use L'Hopital's rule either because the limit is of the form $1^{\infty}$, and I have no clue how to manipulate it so I can apply it. I'm tempted to say that it approaches infinity, but I can't be confident that there isn't another way to solve it.

Any tips? Thanks.

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How can you "change it to" $-2 = (-\frac{2}{e})^n$? It is completely unclear what you mean by that. – Thomas Andrews May 13 '13 at 15:44
@ThomasAndrews I took a guess that I could set $f(n)=2+(-2/e)^n$ and then set $f(n)=0$ and manipulate the equation algebraically. – agent154 May 13 '13 at 15:45
But why would that work even if the limit was zero? What would solving $f(n)=0$ give you? – Thomas Andrews May 13 '13 at 15:51