Evaluate: $\lim_{h \to 0} \int_{-1}^{1}\frac{h}{h^2+x^2}~dx$ 
How can I evaluate: $$\lim_{h \to 0} \int_{-1}^{1}\frac{h}{h^2+x^2}~dx$$

How I proceed:
$$\lim_{h \to 0} \int_{-1}^{1}\frac{h}{h^2+x^2}~dx=2\lim_{h \to 0} \frac{1}{h}\int_{0}^{1}\frac{1}{1+(\frac{x}{h})^2}~dx=2\lim_{h \to 0}\frac{1}{h}\arctan\frac{1}{h}$$ 
Then how can I prooceed. Please help. Thank in advance.

Hint: $$\lim_{x\to +\infty}\arctan x=\frac{\pi}2$$ while
$$\lim_{x\to -\infty}\arctan x=-\frac{\pi}2$$
Therefore,
$$\lim_{h \to 0^+}\frac1h\arctan\frac{1}{h}=\lim_{y \to +\infty}y\arctan y=(+\infty)\frac\pi 2=+\infty$$ 
while
$$\lim_{h \to 0^-}\frac1h\arctan\frac{1}{h}=\lim_{y \to -\infty}y\arctan y=(-\infty)\frac{-\pi} 2=+\infty$$ 
The required limit is
$$\lim_{h \to 0}\frac1h\arctan\frac{1}{h}=+\infty$$ 
