Prove that there exists the limit $\lim\limits_{n \to \infty}\int_0^1\frac{e^x \cos x}{nx^2+\frac{1}{n}}dx$ Use the A-G inequality, we can easily have that $$\left|\frac{e^x \cos x}{nx^2+\frac{1}{n}}\right| \leq \frac{e^x}{2x}$$
however, $\int_0^1 g(x)dx= \int_0^1\frac{e^x}{2x}dx$ diverges ,thus we cannot apply the Lebesgue Dominated Convergence Theorem.\
I have an idea that use the characteristic function $\chi_{[\frac{1}{n},1]}f_n(x)$ to approximate $f_n(x)$. Then $\left|\chi_{[\frac{1}{n},1]}f_n(x)\right| \leq \left|f_n(x)\right| \leq g(x)$ whose integral convergences on $[\frac{1}{n},1]$, then we can apply LDC Theorem to find the limit of $\int_0^1 f_n(x)=0$.\
Is this idea correct or how to write it down rigorously?

By enforcing the substitution $x=\frac{z}{n}$, $dx=\frac{dz}{n}$ we have
$$ \int_{0}^{1}\frac{e^x\cos x}{nx^2+\frac{1}{n}}\,dx = \int_{0}^{n}\frac{e^{z/n}\cos(z/n)}{z^2+1}\,dz=\int_{0}^{+\infty}\frac{f_n(z)}{z^2+1}\,dz -O\left(\frac{1}{n}\right)$$
where $f_n(z)$ is defined as $e^{z/n}\cos(z/n)$ over $[0,n]$ and as $1$ over $[n,+\infty)$.
Here we may easily apply the dominated convergence theorem, since $e^x\cos x$ is a positive and continuous function on $[0,1]$, leading to $0\leq f_n(x)\leq M$. For any $z\in\mathbb{R}^+$ we have $\lim_{n\to +\infty}f_n(z)=1$, hence the wanted limit equals $\int_{0}^{+\infty}\frac{dz}{z^2+1}=\color{red}{\large\frac{\pi}{2}}$.
