A Proof for $\lim_{n \to \infty} (1+\frac{x}{n})^n=e^x$

Question

I found this proof for $\lim_{n \to \infty} (1+\frac{x}{n})^n=e^x$ online and wish to ask one small tiny bit of it

$$e^{\ln{(1 + \frac{x}{n})^n} }=e^{n \ln(1+\frac{x}{n})}$$

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

Apply L'Hopital's Rule:

$$=e^{\lim_{n \to +\infty}\frac{(\frac{-x}{n^2})\frac{1}{1+\frac{x}{n}}}{-\frac{1}{n^2}}} =e^{\lim_{n \to +\infty}\frac{x}{1+\frac{x}{n}}} =e^x$$

What I don't understand is the $\frac{-x}{n^2}$. Where does it come from? Isn't the derivative of $\ln(1+\frac{x}{n})$ is $\dfrac{1}{1+\frac{x}{n}}$ only?

Answer 1

First of all your answer is $e^x$ not $e^n$ so please edit your question.

Secondly derivative of $\ln u $ is $\frac {u'}{u}$ not $1/u$ due to the chain rule of differentiation

Answer 2

In the limit you are differentiating w.r.t to $n$ not with $x$: $$\frac{\mathrm d}{\mathrm dn}\left[\ln\left(1+\frac{x}{n}\right)\right] = \frac{-x}{n^2}.\frac{1}{(1+\frac{x}{n})}$$

Answer 3

I have finally found another proof for this:

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

Now let $h=\frac{x}{n}$, we have $\lim_{h \to 0} e^{x\frac{\ln(1+h)}{h}}$. Since the limit $\frac{\ln(1+h)}{h}=1$, we have arrived at the final result $e^{x}$

Is this way correct?