Why division by infinity is not done immediately in limits I was going through the proof that derivative of $\ln x$ is $\frac{1}{x}$.
Here is what it shows.
$$ \lim_{\Delta x\rightarrow 0}{\frac{\ln\ ( x+\Delta x) -\ln\ x}{\Delta x}}$$
$$
\lim_{\Delta x\rightarrow 0}{\frac{1}{\Delta x} \ln\left(\frac{x+\Delta x}{x}\right)}$$
Let $\displaystyle \Delta x=\frac{x}{n}$ then we have  $$\frac{1}{x} \ln \lim_{n \rightarrow \infty}\left( 1+\displaystyle \frac{1}{n}\right)\displaystyle ^{n}$$
\begin{equation*}
\frac{1}{x} \ \ln\ e=\frac{1}{x}.
\end{equation*}

*

*Why does $n$ does not divide 1 to get 0 giving natural log of 1 raised to power n which will give 0 and the final derivative will be 0.

 A: First, I want to analyze the sentence "$n$ divides $1$ to get $0$ giving natural log of $1$ raised to power $n$ which will give $0$". You appear to suggest the following:
$$
\ln\big(\lim_{n\to\infty}\big(1+\frac{1}{n}\big)^{n}\big)=\ln\big(\big(1+\lim_{n\to\infty}\frac{1}{n}\big)^{n}\big)=\ln((1+0)^n)=\ln(1^{n})=\ln(1)=0
$$
But this has no meaning as I have taken the limiting variable $n$ outside of the limit.
So the correct analysis is to understand that while $n$ approaches infinity, $\frac{1}{n}$ is approaching $0$ and so the base $1+\frac{1}{n}$ is approaching $1$. But at the same time the exponent $n$ is approaching infinity, and so the whole thing is behaving like $1^{\infty}$ in the limit.
However, $1^{\infty}$ is an "indeterminate form" meaning it does not provide sufficient information for determining the limit.
Other indeterminate forms are $\frac{0}{0}$, $\frac{\infty}{\infty}$ and $\infty-\infty$.
For example consider the two limits
$$
\lim_{x\to\infty} \frac{x^2}{x+1}\textrm{ and }\lim_{x\to\infty}\frac{x+1}{x^2}
$$
Both limits look like $\frac{\infty}{\infty}$ but the first limit is $\infty$ and the second limit $0$.
In your case of $1^{\infty}$ the issue is that if $\lim_{x\to\infty}f(x)=1$ and $\lim_{x\to\infty}g(x)=\infty$ then this information alone is not sufficient to determine $\lim_{x\to\infty}f(x)^{g(x)}$. In general, such a limit could take any value from $0$ to $\infty$ inclusive.
As with other indeterminate forms, one can often apply some trick to turn it into a L'Hopital problem. For $1^{\infty}$ the "trick" is via the identity $\exp(\ln t)=t$. So
$$
\lim_{x\to\infty}f(x)^{g(x)}=\lim_{x\to\infty}\exp(\ln(f(x)^{g(x)}))=\exp\big(\lim_{x\to\infty}\ln(f(x)^{g(x)})\big)=\exp\big(\lim_{x\to\infty}g(x)\ln (f(x))\big)
$$
$$
=\exp\big(\lim_{x\to\infty}\frac{\ln(f(x))}{g(x)^{-1}}\big)
$$
We have used here the fact that $\exp(x)$ is continuous so one can interchange it with a limit. Notice that the final limit inside $\exp$ in the last line is of the form $\frac{0}{0}$ since $f(x)\to 1$ and $g(x)\to\infty$. So if you can use something like L'Hopital to evaluate that last limit as some value $L$, then the answer for the original limit will be $\exp(L)$.
Let's do this for your example (probably repeating other answers on this network).
Setting $f(x)=1+\frac{1}{x}$ and $g(x)=x$ the lines above give us
$$
\lim_{x\to\infty}\big(1+\frac{1}{x}\big)^x=\exp\big (\lim_{x\to\infty}\frac{\ln(1+\frac{1}{x})}{x^{-1}}\big)
$$
We do L'Hopital. The numerator is rewritten as $\ln(\frac{x+1}{x})$ and the derivative of this is $\frac{x}{x+1}\cdot \frac{-1}{x^2}$. The derivative of the denominator is $\frac{-1}{x^2}$. So by L'Hopital
$$
\lim_{x\to\infty}\frac{\ln(1+\frac{1}{x})}{x^{-1}}=\lim_{x\to\infty}\frac{\frac{x}{x+1}\cdot \frac{-1}{x^2}}{\frac{-1}{x^2}}=\lim_{x\to\infty}\frac{x}{x+1}=1
$$
So the final answer to the original limit is
$$
\lim_{x\to\infty}\big(1+\frac{1}{x}\big)^x=\exp(1)=e
$$
For more on indeterminate forms see https://en.wikipedia.org/wiki/Indeterminate_form
