Why isn't $\lim \limits_{x\to\infty}\left(1+\frac{1}{x}\right)^{x}$ 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 to your title question. It looks like this argument has not been given yet. Except by robjohn, but as a consequence of a more complicated fact.
If this was true, then we would have in particular
$$
\lim_{n\rightarrow+\infty}\left(1+\frac{1}{n}\right)^n=1
$$
along $\mathbb{N}^*$. But for positive integers, the binomial theorem yields

$$
\left(1+\frac{1}{n}\right)^n=1+n\frac{1}{n}+\frac{n(n-1)}{2}\frac{1}{n^2}+\ldots\geq 1+1=2\qquad\forall n\geq 1.
$$

