Proving $\lim\limits_{n\to\infty}\int_0^{\pi/4} \tan^n{x}\,dx=0$ How would you prove that $\displaystyle\lim\limits_{n\to\infty}\int_0^{\pi/4} \tan^n{x}\,dx=0$.
It is obvious if you see the graph of $\tan^n{x}$ on $(0, \pi/4)$ as $n$ increases but i'm looking for a more algebraic way. 
This result is for connecting the power reduction formula for $\displaystyle\int_0^{\pi/4} \tan^n{x}\,dx$ to Leibniz formula for $\pi$.

Let $0<\varepsilon<\frac{\pi}{4}$. Write the integral
$$I_n=\int_0^{\pi/4}\tan^n x\,\mathrm dx=\int_0^{\pi/4-\varepsilon}\tan^n x\mathrm dx+\int_{\pi/4-\varepsilon}^{\pi/4}\tan^n x\,\mathrm dx=J_n+K_n$$
First notice the limit of $I_n$ exists, because it's decreasing and positive, since $0\le\tan^{n+1}x\le\tan^{n}x$.
Since the function $x\to\tan^n x$ is increasing, the first term $J_n$ is bounded by:
$$J_n\le(\frac{\pi}{4}-\varepsilon)\tan^n(\frac{\pi}{4}-\varepsilon)\le\frac{\pi}{4}\beta^n$$
With $0<\beta<1$.
Since $\tan^n x\le1$, The second term $K_n$ is bounded by:
$$K_n\leq\varepsilon$$
Hence, for all $\varepsilon$, there is a $\beta$ such that $0<\beta<1$ such that for all $n$,
$$0\le I_n\le \frac{\pi}{4}\beta^n+\varepsilon$$
Hence, for all $\varepsilon$,
$$0\le\lim_{n\to\infty} I_n\le\varepsilon$$
So
$$\lim_{n\to\infty} I_n=0$$
