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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
    
@ThomasAndrews I thought I could blindly try to say that since $-2=(-2/e)^n$ then I could apply the rule that $\ln(-2)=n\cdot \ln(-2/e)$ and then try to solve it from there. But that clearly doesn't work for many reasons$ –  agent154 May 13 '13 at 15:53
    
But that wouldn't help, since $\ln -2$ is undefined, but even if you could solve it, what good with it do you to have a solution to $f(n)=0$? –  Thomas Andrews May 13 '13 at 15:54

2 Answers 2

up vote 6 down vote accepted

$e>2$ so $\frac{2}{e}<1$ so
$$\lim_{n \to \infty}{\left(-\frac{2}{e}\right)^n}=0$$ so

$$\lim_{n \to \infty}{2+\left(-\frac{2}{e}\right)^n}=2$$

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Thank you Dear Thomas –  Somaye May 13 '13 at 15:50
    
OK, I see it now -- this is the form of $x^n$ where $0<|x|<1$, so it approaches $0$. Never considered that. Thanks –  agent154 May 13 '13 at 15:51
    
@agent154: You are perhaps looking too much for applicable manipulations and/or theorems, and looking too little at the thing itself. This can be a useful short-term strategy, particularly if you have a good memory. In the long run it causes trouble. Look at the expression, perhaps ask your calculator, $n=10$, $n=100$. Soon you will see what's going on, and why. –  André Nicolas May 13 '13 at 16:37
    
I choose this one. –  Babak S. May 13 '13 at 16:53

What is the absolute value of $2/e$? Is it less than $1$ -- and if so, what does this imply for the second term of the sum?

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$|-2/e|$ is a constant between $0$ and $1$... –  agent154 May 13 '13 at 15:48

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