# Limit at Infinity; Exponential with negative base

I was trying to evaluate the following limit

$$\lim_{n \to \infty}\frac{3+(-1)^n}{n^2}$$

The first thing I tried was L'hospital, but I almost immediately realized that $$\ln(-1)(-1)^x$$ would be the derivative of $$(-1)^n$$, and $$\ln(-1)$$ is not defined over the reals.

Next, I tried the squeeze theorem, but I couldn't think of any bounds for this. Any help would be great.

According to the textbook, the answer is $$0$$

$$0<\frac {3+(-1)^{n}} {n^{2}}<\frac 4 {n^{2}}$$ so the limit is $$0$$.
This sequence can be squeezed between $$\frac{3-1}{4^n}$$ and $$\frac{3+1}{4^n}$$.
Notice that $$\frac{2}{n^{2}}\le\frac{3+\left(-1\right)^{n}}{n^{2}}\le\frac{4}{n^{2}}$$