Limit of n * ln(1+x/n)

Question

How can you compute with the most primitive tools that:

$$ \lim_{\stackrel{n \to \infty}{n > -x}}n \:\ln (1+\frac{x}{n})=x $$

Using l'hospital verifies this. However we hadn't proofed l'hospital at this point.

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The whole proof for context:

For $x \in \mathbb{R}$ is $e^x=\lim \limits_{\stackrel{n \to \infty}{n > -x}}(1+\frac{x}{n})^{n}$.

Without restriction $x \not= 0$. Let $x_n:= 1+\frac{x}{n}$ for $n \in \mathbb{N}$. Then $x_n>0$ for $n >-x$ und $\lim\limits_{n \rightarrow\infty} x_n=1$. Hence $$ 1 \;{=}\; \ln'(1)= \lim_{\stackrel{n \to \infty}{n > -x}}\frac{\ln x_n-\ln 1}{x_n-1}= \frac{1}{x} \lim_{\stackrel{n \to \infty}{n > -x}}n\: \ln(1+\frac{x}{n}), $$ then $$ \lim_{\stackrel{n \to \infty}{n > -x}}n \:\ln (1+\frac{x}{n})=x $$ hence $$ e^x=\lim_{\stackrel{n \to \infty}{n > -x}}e^{n \:\ln (1+\frac{x}{n})}=\lim_{\stackrel{n \to \infty}{n > -x}}(1+\frac{x}{n})^{n} $$

Answer 1

Using the limit definition of $e^x$

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

Answer 2

Hints for you to justify:

$$n\log\left(1+\frac xn\right)=\log\left(1+\frac xn\right)^n\xrightarrow[n\to\infty]{}\log e^x=x$$