$Integrate:\int_{-1}^1 \frac{1}{x^2} \, dx$
Let I be the integral:
$I = \int_{-1}^1 \frac{1}{x^2} \, dx$
At \(x = 0\), \(f(x)\) becomes \(\infty\), so we can split the integral into two parts:
$I = \int_{-1}^0 \frac{1}{x^2} \, dx + \int_{0}^1 \frac{1}{x^2} \, dx$
For the left part:
$\lim_{{h \to 0}^-} \int_{-1}^{-h} \frac{1}{x^2} \, dx$ = $\lim_{{h \to 0}^-} \left[-\frac{1}{x}\right]_{-h}^{-1}$ = $\lim_{{h \to 0}^-} \left[-\frac{1}{-h} - (-1)\right] = \infty$
For the right part:
$\lim_{{h \to 0}^+} \int_{h}^1 \frac{1}{x^2} \, dx $= $\lim_{{h \to 0}^+} \left[-\frac{1}{x}\right]_{h}^1$ = $\lim_{{h \to 0}^+} \left[-\frac{1}{h} - (-1)\right] = \infty$
Since both parts diverge to $\infty$, the overall integral \(I\) diverges:
$I = \infty$
Therefore, the limit as \(h\) approaches 0 does not exist.