Answer
$$\lim\limits_{x\to\infty}\frac{\sin^2x}{x^2+1}=0$$
Work Step by Step
$$\lim\limits_{x\to\infty}\frac{\sin^2 x}{x^2+1}$$
*Strategy: Apply the Squeeze Theorem
1) We see that $$0\le \sin^2x\le1$$
Also, since $x^2\ge0$ for $x\in R$, therefore $x^2+1\gt0$ for $x\in R$
So, $$\frac{0}{x^2+1}\le\frac{\sin^2x}{x^2+1}\le\frac{1}{x^2+1}$$ (the inequality direction remains because $x^2+1\gt0$)
$$0\le\frac{\sin^2x}{x^2+1}\le\frac{1}{x^2+1}\hspace{.5cm}(1)$$
2) We calculate:
- $\lim\limits_{x\to\infty}0=0$
-$\lim\limits_{x\to\infty}\frac{1}{x^2+1}=\frac{1}{\lim\limits_{x\to\infty}(x^2)+1}=\frac{1}{\infty+1}=0$
So, $\lim\limits_{x\to\infty}0=\lim\limits_{x\to\infty}\frac{1}{x^2+1}=0\hspace{.5cm}(2)$
3) From $(1)$ and $(2)$, according to the Squeeze Theorem, we conclude: $$\lim\limits_{x\to\infty}\frac{\sin^2x}{x^2+1}=0$$
