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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 $0$

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3 Answers 3

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$0<\frac {3+(-1)^{n}} {n^{2}}<\frac 4 {n^{2}}$ so the limit is $0$.

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This sequence can be squeezed between $\frac{3-1}{4^n}$ and $\frac{3+1}{4^n}$.

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Notice that $$\frac{2}{n^{2}}\le\frac{3+\left(-1\right)^{n}}{n^{2}}\le\frac{4}{n^{2}}$$

using Squeeze theorem implies the desired result.

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