Answer
$$\lim_{x\to\infty}f(x)=\lim_{x\to-\infty}f(x)=\pi$$
Work Step by Step
$$f(x)=\pi-\frac{2}{x^2}$$
(a) As $x\to\infty$, or $x$ gets infinitely large, $x^2$ will approach $\infty$ and $2/x^2$, hence, will approach $0$.
Therefore, $$\lim_{x\to\infty}f(x)=\lim_{x\to\infty}\Big(\pi-\frac{2}{x^2}\Big)=\pi-0=\pi$$
(b) As $x\to-\infty$, or $x$ gets infinitely small, $x^2$ will approach $\infty$, and $2/x^2$ will approach $0$ as a result.
Therefore, $$\lim_{x\to-\infty}f(x)=\lim_{x\to-\infty}\Big(\pi-\frac{2}{x^2}\Big)=\pi-0=\pi$$
A graph of the function $f(x)$ is enclosed below, which shows that $f(x)$ approaches $\pi$ as $x$ approaches either $\infty$ or $-\infty$.
