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When it comes to find the limit of $(1+x)^{1/x}$ when $x$ goes to infinity,

I put $\frac{1}{x} = t$ and replaced the whole equation with $(1+\frac{1}{t})^t$ when $t$ goes to $0$.

Hence, I wrote the answer as $e$, because I learned that the value approximates to 2.718....

However, when I use the logarithm differentiation, I got $1$.

I got two answer by solving it in two different ways.. What am I doing wrong?

Mary Star
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KeepCoding
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    If the "contest" is between the answers $e$ and $1$, then you can figure out the winner by asking whether the answer should be $>2$ (like $e$) or $<2$ (like $1$). So consider some large $x$, say $1,000,000$, and see whether $(1+x)^{1/x}$ is above or below $2$. Equivalently, is $1+x$ above or below $2^x$? Well, $1,000,001$ is actually below $2^{20}$ (multiply it out and check), so "below" wins by a large margin. The margin only gets more impressive for bigger $x$. So $e$ can't be right. – Andreas Blass Jul 30 '14 at 03:37

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Substituting $t=\frac{1}{x}$ we have $t \to 0$ and not $\infty$

$$\lim_{t \to 0} \left (1+\frac{1}{t}\right )^t \nrightarrow e$$

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

$$\lim_{x \rightarrow \infty} (1+x)^{1/x}=1$$

Mary Star
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