Need to prove $\lim_{n\to\infty} n(log(1+x/n)-x/n) = 0 $

Question

This is the second part of a problem that was on my test.

The first part was:

1. For all $ x > -1$, prove $\log(1+x) - x = \int_{0}^x \frac{(t-x)}{(1+t)^2}dt$

which I did with integration by parts.

The second part was:

2. Assuming part 1, prove part 2 which is:

for all x in R, $\lim_{n\to\infty} n(\log(1+x/n)-x/n) = 0 $

I can see how this works but, I can't seem to prove it. I tried L'hopital's rule which was becoming too messy, then tried working with the formal definition of a sequence but again I was getting stuck.

Please help.

Answer 1

Assuming $x$ is positive, the negative case is similar, $$\int_0^{\frac{x}{n}}\frac{\frac{x}{n}-t}{(1+t)^2}dt\leq \int_0^{\frac{x}{n}}\frac{\frac{x}{n}}{(1+t)^2}dt=\frac{x^2}{n^2}\frac{1}{1+\frac{x}{n}}\to 0$$

Answer 2

Note that

$$n\log(1+x/n)=\log((1+x/n)^n)$$

and we know that $\lim_{n\to\infty}(1+x/n)^n=e^x$

Can you finish it from here?

Answer 3

Note that by Taylor’s expansion

• $\log(1+x/n)=\frac x n-\frac12 \frac {x^2}{n^2}+o(n^{-2})$

then

$$n(\log(1+x/n)-x/n)=-\frac12 \frac {x^2}{n}+o(n^{-1})\to 0$$