Find $\lim_{x\to \infty}{\sin(x^2)}x^{-1/2} $ Find $$\lim_{x\to \infty}\frac{\sin(x^2)}{ \sqrt{x}} $$
(1) $$-1 \le \sin(x^2) \le 1 $$
$$\frac{-1}{\sqrt{x}} \le \frac{\sin(x^2)}{\sqrt{x}} \le \frac{1}{\sqrt{x}}$$
(2) I prove that $\lim \frac{1}{\sqrt{x}} = 0$ using the delta definition  
(3) $$0 \le \lim_{x\to \infty}\frac{\sin(x^2)}{\sqrt{x}} \le 0$$
$$\lim_{x\to \infty}\frac{\sin(x^2)}{\sqrt{x}} =0$$
Have I solved this correctly?
 A: It's not exactly correctly. 
You need to delete $(3)$ because it's unnecessary and 
you need to say also that
$$\lim_{x\rightarrow+\infty}\frac{1}{\sqrt{x}}=\lim_{x\rightarrow+\infty}\left(-\frac{1}{\sqrt{x}}\right)=0$$
A: *

*I believe you mean $\lim \limits_{\color{red}x \to \infty}$

*In $(2)$, write $\lim \limits_{\color{red}{x \to \infty}}\frac{1}{\sqrt{x}}$

*In $(3)$,
Since $$\color{red}{\lim_{x \to \infty}}\frac{-1}{\sqrt{x}} \le \color{red}{\lim_{x \to \infty}}\frac{\sin(x^2)}{\sqrt{x}} \le \color{red}{\lim_{x \to \infty}}\frac{1}{\sqrt{x}}$$
Hence, the conclusion.
Summary: Do not omit the limit.
A: Basiclly the limit is zero, cause the limit lays between two limits and both of them are equal to zero. (it is a theorem)
A: No. You have to state clearly what you know, what the target conclusion is and write a chain of deductions were each step is either trivial or supported by reference to a known theorem.
While used by many textbooks, I must caution you against the application of the $\lim$ notation. The bare statement: 

Find $$\underset{x \rightarrow \infty}{\lim}
\frac{\sin(x^2)}{\sqrt{x}}.$$

obscures the fact that we must accomplish two goals. Specifically, we must show that the limit exists and we must compute the value. Beyond doubt, this point is made once somewhere in every textbook, but important points should be repeated.
Moreover, the notation fails to explicitly identify the range of valid $x$ values.
I will now demonstrate how to avoid these issues completely. Let $f : (0,\infty) \rightarrow \mathbb R$ be given by
$$ f(x) = \frac{\sin(x^2)}{\sqrt{x}}.$$
We must show that
$$ f(x) \rightarrow 0, \quad x \rightarrow \infty, \quad x \in (0, \infty).$$
To that end, we observe that
$$ \forall x \in (0,\infty) \: : \: -\frac{1}{\sqrt{x}} \leq f(x) \leq \frac{1}{\sqrt{x}}.$$
Here we have used $$-1 \leq \sin(x^2) \leq 1,$$ for all $x \in \mathbb R$. Since
$$ -\frac{1}{\sqrt{x}} \rightarrow 0, \quad x \rightarrow \infty, \quad x \in (0,\infty),$$
and
$$ \frac{1}{\sqrt{x}} \rightarrow 0, \quad x \rightarrow \infty, \quad x \in (0,\infty),$$
the Squeeze Lemma allows us to conclude that
$$ f(x) \rightarrow 0, \quad x \rightarrow \infty, \quad x \in (0,\infty).$$
This completes the proof.
