Question on limit: $\lim_{x\to 0}\large \frac{\sin^2{x^{2}}}{x^{2}}$ How would I solve the following trig equations?
$$\lim_{x\to 0}\frac{\sin^2{x^{2}}}{x^{2}}$$
I am thinking the limit would be zero but I am not sure.
 A: $$\lim_{x\to 0} \frac{ \sin^{2} x^2}{x^2} = \lim_{t\to 0} \frac{\sin^2 t }{t}$$ which is equal to $$ \lim_{t \to 0} \frac{\sin t \sin t \cdot t}{t\cdot t} = \lim _{t\to0} \frac{\sin t}{t} \lim _{t\to0} \frac{\sin t}{t} \lim _{t\to 0} t = 1\cdot 1 \cdot 0 =0$$
A: Yet another solution, via the Sandwich theorem: We know that $|\sin x|\leq |x|$ for every $x$. Thus, whenever $x\neq 0$ we have 
$$
0\leq\left|\frac{\sin^2(x^2)}{x^2}\right| =  \frac{\left|\sin(x^2)\right|\cdot\left|\sin(x^2)\right|}{x^2} \leq \frac{x^2\cdot x^2}{x^2}=x^2 .
$$
Now since $\lim_{x\to 0} x^2=\lim_{x\to 0}0=0$, the Sandwich rule gives that $\lim_{x\to 0}\frac{\sin^2(x^2)}{x^2}=0$. 
A: I think you want
$$\lim_{x\to 0}\frac{\sin^2 x^2}{x^2}.$$
Rewrite our function as 
$$\left(\sin x^2\right) \frac{\sin x^2}{x^2}.$$
Now is it easy? Yes, indeed the limit is $0$. 
A: 
We use $\;\sin^2(x^2) = (\sin x^2)(\sin x^2)$
$$\lim_{x\to 0}\frac{\sin^2(x^2)}{x^2}\;=\;\lim_{x \to 0}\; (\sin x^2) \frac{\sin x^2}{x^2} = \lim_{x\to 0} \sin x^2 \cdot 1 = 0$$
Recall, we're also using the fact that $\lim_{t \to 0} \dfrac{\sin t}{t} = 1$. Here, $t = x^2$.

A: We use $$\sin^2 x =\frac{1-\cos 2x}{2}$$
$$\lim_{x\to 0}\frac{\sin^2 x^2}{x^2}.=\lim_{x\to 0} \frac {1-\cos 2x^2}{2x^2}=0  $$
