Finding $\displaystyle \lim_{n \to \infty} \int_0^\infty \frac{e^{-x}\cos{x}}{nx^2 + \frac{1}{n}}dx$ Prove that $\displaystyle \lim_{n \to \infty} \int_0^\infty \frac{e^{-x}\cos{x}}{nx^2 + \frac{1}{n}}dx$ exists and determine its value.
 A: Hint: Break this integral into two parts.
$$ \int_0^\infty \frac{e^{-x}\cos{x}}{nx^2 + \frac{1}{n}}dx=\int_0^1 \frac{e^{-x}\cos{x}}{nx^2 + \frac{1}{n}}dx+\int_1^\infty \frac{e^{-x}\cos{x}}{nx^2 + \frac{1}{n}}dx$$
For the first part we use integration by parts to show that:
$$\int_0^1 \frac{e^{-x}\cos{x}}{nx^2 + \frac{1}{n}}dx=[tan^{-1}(nx)e^{-x}cos(x)]^{1}_{0}-\int_0^1 tan^{-1}(nx)(e^{-x}\cos{x})'dx...(1)$$
The limit of the term $[tan^{-1}(nx)e^{-x}cos(x)]^{1}_{0}$ can be found easily. The limit of the integral $\int_0^1 tan^{-1}(nx)(e^{-x}\cos{x})'dx$ can be found by noting that 
$$\forall x\in[0,1][tan^{-1}(nx)(e^{-x}\cos{x})'\leq\frac{\pi}{4}(e^{-x}\cos{x})']$$
Finally, use Lesbesuge's dominated convergence to find the limit of the last integral apperaring in (1)
The limit of the second part can be found by noting that $\forall n\in Z^+\forall x\geq 1[x^2+1\leq nx^2+\frac{1}{n}]$. Thus:
$$\forall n\in Z^+\forall x\geq1[\frac{e^{-x}\cos{x}}{nx^2 + \frac{1}{n}}\leq\frac{e^{-x}\cos{x}}{x^2 + 1}]$$
Now you can use Lebesuge's dominated convergence theorem to evaluate the second integral
A: Hint: Make the substitution $x=y/n$ and apply the dominated convergence theorem to see that
$$
\lim_{n \to \infty} \int_0^{\infty} \frac{e^{-x} \cos x}{nx^2+1/n}\,dx = \int_{0}^{\infty} \frac{dy}{y^2+1} = \frac{\pi}{2}.
$$
