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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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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. –  joriki Mar 19 '13 at 1:55
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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. –  DonAntonio Mar 19 '13 at 2:12

2 Answers 2

up vote 10 down vote accepted

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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Very nicely done +1 –  amWhy Mar 19 '13 at 2:06

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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Why can you interchange $\lim$ and $()^{-1}$? –  newbie Mar 19 '13 at 2:41
    
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 $\,(a^n)^m=a^{nm}\,$...:) –  DonAntonio Mar 19 '13 at 2:52

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