Question

The value of the $\lim_{n\to\infty} n\int_{0}^{\pi/4} \tan^n x \,dx$ is

Answer 1

1/2

Answer 2

To evaluate the limit $\lim_{n\to\infty} n\int_{0}^{\pi/4} \tan^n x \,dx$, let $I_n = \int_{0}^{\pi/4} \tan^n x \,dx$.

Step 1: Derive a recurrence relation for $I_n$.

We can relate $I_n$ and $I_{n+2}$ using integration by parts or by manipulating the integrand.
Consider $I_{n+2}$:

$I_{n+2} = \int_{0}^{\pi/4} \tan^{n+2} x \,dx = \int_{0}^{\pi/4} \tan^n x \cdot \tan^2 x \,dx$

We know that $\tan^2 x = \sec^2 x - 1$. Substitute this into the integral:

$I_{n+2} = \int_{0}^{\pi/4} \tan^n x (\sec^2 x - 1) \,dx$ $I_{n+2} = \int_{0}^{\pi/4} \tan^n x \sec^2 x \,dx - \int_{0}^{\pi/4} \tan^n x \,dx$

The second term on the right is $I_n$.
For the first term, let $u = \tan x$. Then $du = \sec^2 x \,dx$.
When $x=0$, $u=\tan 0 = 0$. When $x=\pi/4$, $u=\tan(\pi/4) = 1$.
So, the first integral becomes:

$\int_{0}^{1} u^n \,du = \left[ \frac{u^{n+1}}{n+1} \right]_{0}^{1} = \frac{1^{n+1}}{n+1} - \frac{0^{n+1}}{n+1} = \frac{1}{n+1}$

Therefore, the recurrence relation is:

$I_{n+2} = \frac{1}{n+1} - I_n$

Rearranging this, we get:

$I_n + I_{n+2} = \frac{1}{n+1}$

Step 2: Show that $\lim_{n\to\infty} I_n = 0$.

For $x \in [0, \pi/4)$, we have $0 \le \tan x < 1$. At $x=\pi/4$, $\tan x = 1$.
Let $\delta$ be a small positive number. We can split the integral:

$I_n = \int_{0}^{\pi/4 - \delta} \tan^n x \,dx + \int_{\pi/4 - \delta}^{\pi/4} \tan^n x \,dx$

For the first integral, since $0 \le x \le \pi/4 - \delta$, we have $0 \le \tan x \le \tan(\pi/4 - \delta)$. Let $k = \tan(\pi/4 - \delta)$. Since $\pi/4 - \delta < \pi/4$, we have $k < 1$.
So, $\int_{0}^{\pi/4 - \delta} \tan^n x \,dx \le \int_{0}^{\pi/4 - \delta} k^n \,dx = (\pi/4 - \delta) k^n$.
As $n \to \infty$, since $k < 1$, $k^n \to 0$. Thus, $\lim_{n\to\infty} \int_{0}^{\pi/4 - \delta} \tan^n x \,dx = 0$.
For the second integral, since $\pi/4 - \delta \le x \le \pi/4$, we have $0 \le \tan x \le 1$.
So, $0 \le \int_{\pi/4 - \delta}^{\pi/4} \tan^n x \,dx \le \int_{\pi/4 - \delta}^{\pi/4} 1^n \,dx = \int_{\pi/4 - \delta}^{\pi/4} 1 \,dx = \pi/4 - (\pi/4 - \delta) = \delta$.
Since $\delta$ can be chosen arbitrarily small, this implies that $\lim_{n\to\infty} \int_{\pi/4 - \delta}^{\pi/4} \tan^n x \,dx = 0$.
Therefore, $\lim_{n\to\infty} I_n = 0$.

Step 3: Evaluate the limit.

We have the recurrence relation $I_n + I_{n+2} = \frac{1}{n+1}$.
Multiply the entire equation by $n$:

$nI_n + nI_{n+2} = \frac{n}{n+1}$

Now, take the limit as $n \to \infty$:

$\lim_{n\to\infty} (nI_n + nI_{n+2}) = \lim_{n\to\infty} \frac{n}{n+1}$

Let $L = \lim_{n\to\infty} nI_n$.
The right-hand side limit is:

$\lim_{n\to\infty} \frac{n}{n+1} = \lim_{n\to\infty} \frac{1}{1 + 1/n} = \frac{1}{1+0} = 1$

For the left-hand side, we have $\lim_{n\to\infty} nI_n = L$.
For the second term, $\lim_{n\to\infty} nI_{n+2}$:

$nI_{n+2} = \frac{n}{n+2} \cdot (n+2)I_{n+2}$

As $n \to \infty$, $\lim_{n\to\infty} \frac{n}{n+2} = \lim_{n\to\infty} \frac{1}{1 + 2/n} = 1$.
Also, $\lim_{n\to\infty} (n+2)I_{n+2} = L$ (since $n+2$ is just a shifted index, the limit value remains the same).
So, $\lim_{n\to\infty} nI_{n+2} = 1 \cdot L = L$.
Substituting these limits back into the equation:

$L + L = 1$ $2L = 1$ $L = \frac{1}{2}$

The value of the limit is $1/2$.