How to compute $\lim_{n\to \infty}\int_{0}^{n}\frac{\cos(\frac{\pi t}{n})-e^{-t^2}}{n^2 \arcsin(\frac{t}{n})\ln(1+\frac{t}{n})}dt$? If $c$ is a constant, then we can compute $\lim_{n\to \infty}\int_{0}^{c}\frac{\cos(\frac{\pi t}{n})-e^{-t^2}}{n^2 \arcsin(\frac{t}{n})\ln(1+\frac{t}{n})}dt$ by Lebesgue dominated convergence theorem. But for $\lim_{n\to \infty}\int_{0}^{n}\frac{\cos(\frac{\pi t}{n})-e^{-t^2}}{n^2 \arcsin(\frac{t}{n})\ln(1+\frac{t}{n})}dt$, the upper limit of integral $\int_{0}^{n}\frac{\cos(\frac{\pi t}{n})-e^{-t^2}}{n^2 \arcsin(\frac{t}{n})\ln(1+\frac{t}{n})}dt$ is $n$, so I don't know how to solve it.
 A: Let's denote $\displaystyle I(n)=\int_0^n\frac{\cos(\frac{\pi t}{n})-e^{t^2}}{n^2 \arcsin(\frac{t}{n})\ln(1+\frac{t}{n})}dt$.
Making the substitution $t=nx$ and splitting the interval of integration $\displaystyle [0;\epsilon];\,[\epsilon;1]$
$$I(n)=\frac{1}{n}\int_0^\epsilon\frac{\cos\pi x-e^{-n^2x^2}}{\arcsin x \ln(1+x)}dx\,+\,\frac{1}{n}\int_\epsilon^1\frac{\cos\pi x-e^{-n^2x^2}}{\arcsin x \ln(1+x)}dx=I_1+I_2$$
$$\Big|I_2\Big|<\frac{1}{n}\int_\epsilon^1\frac{\cos\pi \epsilon+e^{-n^2\epsilon^2}}{\arcsin \epsilon \,\ln(1+\epsilon)}dx=\frac{(1-\epsilon)}{n}\frac{\cos\pi \epsilon+e^{-n^2}}{\arcsin \epsilon \ln(1+\epsilon)}\to 0\,\,\text{at}\,\, n\to\infty\,\,\text{for a fixed }\,\epsilon$$
To evaluate $I_1$ we notice that
$$\arcsin x=x+\frac{1}{2}\,\frac{x^3}{3}+\frac{1\cdot 3}{2\cdot4}\,\frac{x^5}{5}+...$$
so both $\arcsin x$ and $\frac{\arcsin x}{x}$ are growing functions, and that $\frac{\arcsin x}{x}\Big|_{x=0}=1$. Therefore, for $x\in[0;\epsilon]$
$$1\leqslant\frac{\arcsin x}{x}\leqslant\frac{\arcsin \epsilon}{\epsilon}\,\,\Rightarrow\,\,x\leqslant\arcsin x\leqslant\frac{\arcsin \epsilon}{\epsilon}\,x\tag{1}$$
In the similar way, we use the fact that $\frac{\ln(1+x)}{x}$ is a declining function for $x\in[0;1]$, and that $\,\frac{\ln(1+x)}{x}\Big|_{x=0}=1$
$$\frac{\ln(1+\epsilon)}{\epsilon}\leqslant\frac{\ln(1+x)}{x}\leqslant 1\,\,\Rightarrow\,\,\frac{\ln(1+\epsilon)}{\epsilon}\,x\leqslant\ln(1+x)\leqslant x\tag{2}$$
Using (1) and (2), we can evaluate $I_1$ as
$$\frac{1}{n}\frac{\epsilon}{\arcsin\epsilon}\int_0^\epsilon\frac{\cos\pi x-e^{-n^2x^2}}{x^2}dx\leqslant\frac{1}{n}\int_0^\epsilon\frac{\cos\pi x-e^{-n^2x^2}}{\arcsin x \ln(1+x)}dx$$
$$\leqslant\frac{1}{n}\frac{\epsilon}{\ln(1+\epsilon)}\int_0^\epsilon\frac{\cos\pi x-e^{-n^2x^2}}{x^2}dx $$
Integrating by part both sides (for example, LHS;  also, expanding integration to $\infty$)
$$\frac{1}{n}\int_0^\epsilon\frac{\cos\pi x -e^{-n^2x^2}}{x^2}dx=-\frac{1}{n}\frac{(\cos\pi x-e^{-n^2x^2})}{x}\bigg|_{x=0}^\epsilon+\,\frac{1}{n}\int_0^\infty \frac{2n^2x\,e^{-n^2x^2}-\pi\sin\pi x}{x}dx\,$$
$$-\,\frac{1}{n}\int_\epsilon^\infty \frac{2n^2x\,e^{-n^2x^2}-\pi\sin\pi x}{x}dx\tag{3}$$
Integrating, making the estimation of every term and taking the limit $n\to\infty$ in (3), we get $\sqrt \pi$, and the following inequality holding:
$$\frac{\epsilon}{\arcsin\epsilon}\sqrt\pi\leqslant \lim_{n\to\infty}I_1\leqslant \frac{\epsilon}{\ln(1+\epsilon)}\sqrt \pi$$
Because $\epsilon\in (0;1)$ is an arbitrary small constant, we conclude that $$\lim_{n\to\infty}I(n)=\lim_{n\to\infty}\Big(I_1+I_2\Big)=\sqrt \pi$$
