$$\lim_{x\to \infty} \frac{x^2(1+\sin^2x)}{(x+\sin x)^2}$$
I can't figure out how to manipulate this algera so as to get the limit I want. Any hint?
Answer
The limit doesn't exist. Note that, dividing by $x^2$ the numerator and denominator, we arrive at the equality
$$\lim_{x\to \infty} \frac{x^2(1+\sin^2 x)}{(x+\sin x)^2}=\lim_{x\to \infty} \frac{1+\sin^2x}{\left(1+\frac{\sin x}{x}\right)^2}.$$ But, since $\left(1+\frac{\sin x}{x}\right)^2\to 1$ as $x\to \infty$ and $\lim_{x\to \infty}(1+\sin^2x)$ doesn't exist, we conclude that the original one doesn't exist.
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