Given $|x|<1$, prove $\lim\limits_{n \to \infty} \int\limits_0^x t^n/(1+t)\ dt = 0$

Question

I have to prove that if |x|<1 , then

$$\lim_{n \to \infty}\displaystyle \int_0^{x} \frac{(-t)^n}{1+t} dt$$ is 0. As this integral has no primitive, I am not sure how to do this. Can you give me some help with this?

Thanks!

Answer 1

$$\left|\int_0^{x} \frac{(-t)^n}{1+t} dt \right| \le \int_{[0,x]} \frac{|x|^n}{1-|x|} dt \le \frac{|x|^{n+1}}{1-|x|} \to 0$$

Answer 2

It's actually true for $x=1$ as well:

$$\left | \int_0^{1} \frac{(-t)^n}{1+t} dt\right | \le \int_0^{1} \left |\frac{(-t)^n}{1+t}\right| dt = \int_0^{1} \frac{t^n}{1+t}\, dt < \int_0^{1} t^n\, dt = \frac{1}{n+1} \to 0.$$