How can I solve the limits: $\mathop {\lim }\limits_{x \to \infty } {{\sin {x^2}} \over {{x^2} + x}}$

Question

When I use $\sin x \sim x$ , the answer is $1$ , is the answer correct?

Answer 1

No. $|sinx^2| \leq 1$, then apply squeeze theorem, you get 0

And you can only use $x$ to approximate $sinx$ when $x$ is close to 0. This is because $\lim_{x\to 0} \frac{sinx}{x}=1$. It's not true when $x$ is large.

Answer 2

Hint: $$-\dfrac{1}{x^2+x}\leqslant\dfrac{\sin x^2}{x^2+x}\leqslant\dfrac{1}{x^2+x},\tag{$x\gt0$}$$ then squeeze everything.

Answer 3

We can use $sinx \sim x$ only when $x \to 0$ , when $x \to \infty $, we cannot use this substitution.

We may solve this problem in this way: $$0 \leftarrow - \mathop {\lim }\limits_{x \to \infty } {1 \over {{x^2} + x}} \le \mathop {\lim }\limits_{x \to \infty } {{\sin {x^2}} \over {{x^2} + x}} \le \mathop {\lim }\limits_{x \to \infty } {1 \over {{x^2} + x}} \to 0$$

So $$\mathop {\lim }\limits_{x \to \infty } {{\sin {x^2}} \over {{x^2} + x}} = 0$$

If any questions, let me know please.

Answer 4

in the limit $x \to \infty$, the numerator is bounded between $[-1,1]$ while the denominator approach infinity so the limit must be zero.