Need to find $\lim_{n \to \infty} n \log({1+x/n})$

Question

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

Answer 1

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 $|y|<1$

Answer 2

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 $z\mapsto\log(1+z)$.

Answer 3

$$ \lim_{n\to\infty} n\log\left(1+\frac xn\right) = x\lim_{1/n\to0} \frac{\log(1+x/n)-\log 1}{x/n} $$

$$ = x\lim_{h\to0}\frac{\log(1+h)-\log1}h = x\cdot\left.\frac{d}{du}\log u\right|_{u=1}. $$