TITLE: Show the limit is $1$ using L'Hòpital rule
QUESTION [1 upvotes]: $\lim_{n \to \infty} n\ln\left(1+\frac{1}{n}\right)$ 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.

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