Compute $\lim_{n\to\infty} \int_{0}^{\pi/4}\tan^n x \ dx$

Use the substitution $t:=\tan x$ (then $\arctan t=x$ and $dx=\frac{dt}{1+t^2}$). Then $$0\leq I_n:=\int_0^{\frac{\pi}4}\tan^n xdx=\int_0^1\frac{t^n}{1+t^2}dt\leq \int_0^1t^ndt=\left[\dfrac{t^{n+1}}{n+1}\right]_{0}^{1}=\frac 1{n+1},$$ which gives $0$ as limit.

The formula also gives the recursion relation $I_{n+2}=\frac 1{n+1}-I_n$, which can help to study the asymptotic behavior of $\{I_n\}$.

$$I_n = \int_0^{\pi/4} \tan^n(x) dx = \int_0^{\pi/4} \tan^{n-2}(x) \sec^2(x) dx - \int_0^{\pi/4} \tan^{n-2}(x)dx$$ $$I_n + I_{n-2}= \int_0^1 t^{n-2} dt = \dfrac1{n-1}$$ Note that $I_n$ is monotone decreasing with $n$ and is bounded below by $0$ and hence it converges.

Hence, if $\lim_{n \to \infty} I_n = L$, we have that $L + L = \lim_{n \to \infty} \dfrac1{n-1} \implies L = 0$

Use the Lebesgue Dominated Convergence Theorem, or the Monotone Convergence Theorem.

EDIT: Or explicitly, given $\pi/4 > \epsilon > 0$,

$$\int_0^{\pi/4} \tan^n x\ dx < (\pi/4 - \epsilon) \tan(\pi/4 - \epsilon)^n + \epsilon $$ which is less than $2 \epsilon$ if $n$ is large enough.