Get the solution to "Solving limits: f(x)=limn→∞(n∫0π4(tan⁡x)ndx)"

Cynthia Bell

Answered question

2022-01-04

Solving limits:

f (x) = lim_{n \to \infty} (n \int_{0}^{\frac{π}{4}} {(\tan x)}^{n} d x)

Answer & Explanation

Hector Roberts

Beginner2022-01-05Added 31 answers

You may perform the change of variable

u = \tan x

to get easily

I_{n} = \int_{0}^{\frac{π}{4}} {(\tan x)}^{n} d x = \int_{0}^{1} \frac{u^{n}}{1 + u^{2}} d u

Then you may just integrate by parts,

I_{n} = \int_{0}^{1} \frac{u^{n}}{1 + u^{2}} d u = \frac{u^{n + 1}}{(n + 1)} \frac{1}{1 + u^{2}} ∣_{0}^{1} + \frac{2}{(n + 1)} \int_{0}^{1} \frac{u^{n + 2}}{{(1 + u^{2})}^{2}} d u

= \frac{1}{2} \frac{1}{(n + 1)} + \frac{2}{(n + 1)} \int_{0}^{1} \frac{u^{n + 2}}{{(1 + u^{2})}^{2}} d u

(1)
Observing that

0 \leq \int_{0}^{1} \frac{u^{n + 2}}{{(1 + u^{2})}^{2}} d u \leq \int_{0}^{1} u^{n} d u = \frac{1}{n + 1}

gives

0 \leq \frac{2}{(n + 1)} \int_{0}^{1} \frac{u^{n}}{{(1 + u^{2})}^{2}} d u \leq \frac{2}{{(n + 1)}^{2}}

(2)
Then combining (1) and (2) leads to

lim_{n \to + \infty} n \int_{0}^{\frac{π}{4}} {(\tan x)}^{n} d x = lim_{n \to + \infty} n I_{n} = \frac{12}{}

Orlando Paz

Beginner2022-01-06Added 42 answers

Heres

karton

Expert2022-01-11Added 613 answers

After letting $\tan (x) \mapsto x$ and using the elementary limit $lim_{n \to \infty} n \int_{0}^{1} x^{n} f (x) d x = f (1)$ where f(x) is continuous, we conclude that
$lim_{n \to \infty} (n \int_{0}^{π / 4} (\tan x)^{n} d x) = \frac{1}{2}$

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