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While looking through an example in Carothers' Real Analysis, I came across the following:

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

and of course I noticed that it looks similar to $e=\lim_{n\to\infty}\left(1+\frac{1}{n}\right)^n$, so since I didn't see how to get the first expression right away, I tried to prove it.

Since $e=\lim_{n\to\infty}\left(1+\frac{1}{n}\right)^n$, if I take the $x^{\text{th}}$ power on both sides I would get $$e^x=\left(\lim_{n\to\infty}\left(1+\frac{1}{n}\right)^n\right)^x$$

I'm not sure how to simplify this further. My attempt:

Assuming $\left(\lim_{n\to\infty}\left(1+\frac{1}{n}\right)^n\right)^x=\lim_{n\to\infty}\left(1+\frac{1}{n}\right)^{nx}$,

(Fairly certain this assumption is incorrect though..)

I then make the change of variable $k=nx$. Also note that $k=nx\implies n=\frac{k}{x}$

Substituting these in, I get that

$$\lim_{k\to\infty}\left(1+\dfrac{1}{\frac{k}{x}}\right)^k=\lim_{k\to\infty}\left(1+\frac{x}{k}\right)^k$$

Hence $$e^x=\lim_{k\to\infty}\left(1+\frac{x}{k}\right)^k$$

My question can essentially be summarized as follows:

Is this valid: $$\left(\lim_{n\to\infty}\left(1+\frac{1}{n}\right)^n\right)^x=\lim_{n\to\infty}\left(1+\frac{1}{n}\right)^{nx}$$

If not, how would you go about trying to prove the initial statement?

Community
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Sujaan Kunalan
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2 Answers2

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Exponentiation is continuous, and therefore commutes with taking limits. So yes.

Arthur
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It suffices to check whether $$\lim_{y\to y_0} y^x = y_0^x$$ where $y_0>0$. Note that $x$ is fixed, so the above can be written as $$ \lim_{y\to y_0} e^{ x\ln y}= e^{x\ln y_0}$$ Now the claim is obvious by the continuity of the exponential and logarithmic function.

Dimitris
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