Need to find $\lim_{n \to \infty} n \log({1+x/n})$ As part of another problem I have reduced it to the point that I need to find the following limit: 
$\lim_{n \to \infty} n \log({1+x/n})$
But I don't know how to do this one.
Ted
 A: Putting $h=\frac1n,$
Method $1:$ 
$$\lim_{n\to\infty}n\ln\left(1+\frac xn\right)=\lim_{h\to0}\frac{\ln(1+xh)}h$$ which is of the form $\frac 00$
So, applying L'Hospitals' Rule $$\lim_{h\to0}\frac{\ln(1+xh)}h=\lim_{h\to0}\frac x{1+xh}=x$$
Method  $2:$ 
We know,  $$\lim_{y\to0}\frac{\ln(1+y)}y=1$$
Method  $3:$ 
Using Taylor's Expansion, $\ln(1+y)=y+\frac{y^2}2+\frac{y^3}3+\cdots$  for $|y|<1$
A: Think of $n$ as taking real values instead of integer values.  Then this is a L'Hopital's rule problem!  You can write
$$
\lim_{n\rightarrow\infty}n\log\left(1+\frac{x}{n}\right)=\lim_{n\rightarrow\infty}\frac{\log\left(1+\frac{x}{n}\right)}{\frac{1}{n}},
$$
and note here that the numerator and denominator both approach $0$ as $n\rightarrow\infty$ (so that L'Hopital's rule applies.)
You could also note that if $y_n=n\log\left(1+\frac{x}{n}\right)$, then
$$
e^{y_n}=e^{n\log(1+\frac{x}{n})}=\left(1+\frac{x}{n}\right)^n,
$$
and this may be a limit that you've seen before.  You could finish by concluding that because $z\mapsto \ln(z)$ is continuous, and $e^{y_n}\rightarrow L$, it must be the case that $y_n=\ln(e^{y_n})\rightarrow\ln(L)$ as $n\rightarrow\infty$.
Alternatively, if you've seen Taylor Series before, you could make use of the Taylor series for the function $z\mapsto\log(1+z)$.
A: $$
\lim_{n\to\infty} n\log\left(1+\frac xn\right) = x\lim_{1/n\to0} \frac{\log(1+x/n)-\log 1}{x/n}
$$
$$
= x\lim_{h\to0}\frac{\log(1+h)-\log1}h = x\cdot\left.\frac{d}{du}\log u\right|_{u=1}.
$$
