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My question is how to argue the following statement

$$\lim_{n\rightarrow\infty}\left(1-\frac{x}{n}\right)^{-n} = e^{x}.$$

My solution is using the binomial series of $\left(1-\frac{x}{n}\right)^{-n}$ followed by taking the limit and finally converting back into $e^{-x}$.

I'm wondering if there is a more straightforward way to prove this, saying only limit computations.

And my definition of exponential function is given as following $$\lim_{n\rightarrow\infty}\left(1+\frac{x}{n}\right)^{n} = e^{x}$$

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    $\begingroup$ This is false. For $x$ positive, the base is less than $1$, so a negative power of it is greater than $1$, whereas the alleged limit is less than $1$. Note also that the title doesn't reflect the body of the question; according to the body, you're looking for the limit, not for a limit of the limit. $\endgroup$ – joriki Mar 19 '13 at 1:55
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    $\begingroup$ As already commented in one of the answers below, the required limit must have $\,e^x\,$ in the right hand and not $\,e^{-x}\,$ , @newbie. $\endgroup$ – DonAntonio Mar 19 '13 at 2:12
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If we already have the basic properties of the exponential function, we can instead calculate $$\lim_{t\to 0^+} (1-tx)^{-1/t}.$$ To compute this, note that the logarithm of our expression is $$-\frac{\log(1-xt)}{t}.$$ One round of L'Hospital's Rule will find the limit of this. The result is $x$, so the limit of your expression is $e^x$.

Remark: It is $e^x$, not $e^{-x}$.

If we have defined $e^w$ as the limit of $(1+w/n)^n$, then essentially no calculation is needed, since $\left(1-\frac{x}{n}\right)^{-n}=\frac{1}{\left(1-\frac{x}{n}\right)^{n}}$.

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    $\begingroup$ Very nicely done +1 $\endgroup$ – Namaste Mar 19 '13 at 2:06
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An idea using what you know and a little arithmetic of limits:

$$\lim_{n\to\infty}\left(1-\frac{x}{n}\right)^{-n}=\lim_{n\to\infty}\left[\left(1+\frac{(-x)}{n}\right)^n\right]^{-1}=\left(e^{-x}\right)^{-1}=e^x$$

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  • $\begingroup$ Why can you interchange $\lim$ and $()^{-1}$? $\endgroup$ – newbie Mar 19 '13 at 2:41
  • $\begingroup$ I didn't do that: all is within the range of that limit. I just used the algebraic equality $\,a^{-1}=1/a\,$ and, of course, that $$\lim a_n=a\neq 0\iff \lim\frac{1}{a_n}=\frac{1}{a}$$ Well, I also used $\,(a^n)^m=a^{nm}\,$...:) $\endgroup$ – DonAntonio Mar 19 '13 at 2:52
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While other answers have assumed $e^{x} e^{-x} =1$ this does require an effort to prove if we use the definition of $e^{x} $ given in the question and is in fact the key to the solution. We can easily note that if $|x|<n$ then by Bernoulli's inequality we have $$1-\frac{x^{2}}{n}\leq \left(1-\frac{x^{2}}{n^{2}}\right)^{n}\leq 1$$ and hence we have via Squeeze theorem $$\lim_{n\to\infty} \left(1+\frac{x}{n}\right)^{n}\left(1-\frac{x}{n}\right)^{n}=1$$ and this means that $$\lim_{n\to\infty} \left(1-\frac{x}{n}\right)^{-n}=\lim_{n\to\infty}\left(1+\frac{x}{n}\right)^{n}=e^{x}$$

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