Show the limit is $1$ using L'Hòpital rule

Question

$\lim_{n \to \infty} n\ln\left(1+\frac{1}{n}\right)$ using L'Hòpital rule show that this is $1$. Can you do this since there isn't a division and $n$ will obviously tend to infinity and $\ln\left(1+\frac{1}{n}\right)$ will tend to $0$? So there limits aren't matching?

So I set $u=n $

$du=1$

$v= \ln\left(1+\frac{1}{n}\right)$

$dv= -\frac{1}{n^2+n}$

Hence $\ln\left(1+\frac{1}{n}\right) - \frac{1}{n+1}$ in which both of these tend to $0$ so I am complete lost.

Answer 1

Hint

$$\lim_{n\to \infty} n\ln \left(1+\frac1n\right)=\lim_{n\to \infty} \frac{\ln \left(1+\frac1n\right)}{\frac 1n}.$$

Answer 2

$$\lim_{n\to \infty}n\log\left(1+\frac1n\right)=\lim_{n\to \infty}\frac{\log\left(1+\frac1n\right)}{1/n}\lim_{n\to \infty}\frac{\left(\frac{1}{1+1/n}\right)(-1/n^2)}{-1/n^2}=1$$

Answer 3

Using L'Hospital's rule here is properly ridiculous: set $\;\dfrac1n=x$ and observe we have a rate of variation: $$\lim_{n\to\infty}n\ln\Bigl(1+\frac1n\Bigr)=\lim_{x\to0}\frac{\ln(1+x)}x=\bigl(\ln(1+x)\bigr)'_{x=0}=1.$$