# Find limit with integrals $\lim_{n\to \infty} n(\sqrt[n^2]{nI_n}-1)$

Given that

$$I_n=\int_0^{\pi/4}e^{nx}(\tan^{n-1}x+\tan^nx+\tan^{n+1}x)dx,$$

I have to find the limit

$$\lim_{n\to \infty} n(\sqrt[n^2]{nI_n}-1)$$

I tried to use $$0 < \text{tg} x < 1$$ for $$0 and i found that $$0. I think with this, the limit should be $$0$$, but I don't know.

• Welcome to Math.SE! Interesting idea, but with $n\to \infty$ you have $e^{nx}/n \to \infty$ as well – gt6989b Feb 25 '20 at 22:22

Split $$I_n=I_1+I_2$$, where:
$$I_1=\int_0^{\frac{\pi}{4}}e^{nx}(\tan^{n-1}x+\tan^{n+1}x)\,dx,\ \ \ I_2=\int_0^{\frac{\pi}{4}}e^{nx}(\tan^{n}x)\, dx$$
\begin{aligned} I_1&=\int_0^{\frac{\pi}{4}}e^{nx}\tan^{n-1}x (1+\tan^2x)\,dx\\ &= \int_0^{\frac{\pi}{4}}e^{nx}\tan^{n-1}x\cdot (\tan x)'\,dx\\ &= \int_0^{\frac{\pi}{4}}e^{nx}\frac{1}{n}(\tan^n x)'\,dx\\ &= \frac{1}{n}e^{nx}\tan^nx\bigg|_0^{\frac{\pi}{4}} - \frac{1}{n}\int_0^{\frac{\pi}{4}}ne^{nx}(\tan^{n}x)\, dx\\ &=\frac{1}{n} e^{\frac{n\pi}{4}}-I_2 \end{aligned}
Thus $$I_n=\dfrac{1}{n} e^{\frac{n\pi}{4}}$$, and the limit is:
\begin{aligned} \lim_{n\to \infty} n(\sqrt[n^2]{nI_n}-1) &= \lim_{n\to \infty}n(e^{\frac{\pi}{4n}}-1) \\ &= \frac{\pi}{4} \cdot \lim_{n\to \infty}\frac{e^{\frac{\pi}{4n}}-1}{\frac{\pi}{4n}}\\ &=\frac{\pi}{4}\\ \end{aligned}