limits - Why isn't $lim limits_{xtoinfty}left(1+frac{1}{x}right)^{x}$ equal to $1$?

Given $\lim \limits_{x\to\infty}(1+\frac{1}{x})^{x}$, why can't you reduce it to $\lim \limits_{x\to\infty}(1+0)^{x}$, making the result "$1$"? Obviously, it's wrong, as the true value is $e$. Is it because the $\frac{1}{x}$ is still something even though it's really small? Then why is

$$\lim_{x\to\infty}\left(\frac{1}{x}\right) = 0\text{?}$$

What is the proper way of calculating the limit in this case?

Answer

Let $f(x,y)=(1+y)^x$. True enough, $f(x,0)=1$ for every $x$ but this is irrelevant to the limit of $f(x,1/x)$ when $x\to+\infty$. Note that one could also consider $f(\infty,1/x)=\infty$ for every positive $x$, as irrelevant as the preceding value $1$.

To compute the actual limit of $f(x,1/x)$, several approaches exist. One is to look at $\log f(x,1/x)=x\log(1+1/x)$ and to remember that $\log(1+u)\sim u$ when $u\to0$ hence $\log f(x,1/x)\to1$ and $f(x,1/x)\to\mathrm e$.

To see why $\log(1+u)\sim u$ when $u\to0$, consider $g(u)=\log(1+u)$ and note that $g(0)=0$ while $g'(u)=1/(1+u)$ hence $g'(0)=1$ and the Taylor expansion $g(u)=g(0)+g'(0)u+o(u)$ yields the result.

Finally, note that, for every fixed $c$, $f(x,c/x)=(1+c/x)^x\to\mathrm e^c$ hence one can realize every positive limit $\mathrm e^c$ by considering the regimes $x\to+\infty$, $xy\to c$. The limit $1$ is realized if $x\to+\infty$ while $xy\to0$ and the limit $+\infty$ if $x\to+\infty$ while $xy\to+\infty$.