Finding limit of $\sin(x^2/2)/\sqrt{2}\sin^2(x/2)$ as $x\rightarrow0$

Question

Can anybody help me find the limit as $x$ tends to $0$, for $$\frac{\sin(x^2/2)}{\sqrt{2}\sin^2(x/2)}.$$ How can I simplify the expression or use equivalent transformations to find a limit (without using L'Hospital)?

Answer 1

I will be using just one fact to solve this, and that is,

$$\lim_{x\to 0} \frac{ \sin x}{x} = 1$$

So lets start! $$\lim_{x\to 0} \frac {\sin \frac {x^2}{2}}{\sqrt{2} \; \sin^2 \frac {x}{2}}$$

Now, Multiply and divide by, $\frac{x^2}{2} \;$ and $(\frac {x}{2})^2$

The $\frac{x^2}{2}$ term in the denominator along with $\sin \frac {x^2}{2}$ in the numerator tends to ONE. And so is the case with the $(\frac {x}{2})^2$ in the numerator and $\sin^2 \frac {x}{2}$ in the denominator.

After that we see that the remaining $x^2$ terms cancel out (because we have one both in the numerator and the denominator).

Resulting in the answer (cue the drumroll) $$\sqrt 2$$

Edit: I have included the steps taken just as a measure

Answer 2

As $\dfrac{\sin x}x=1$, in multiplicative expressions you can replace $\sin x$ by $x$.

Hence

$$\lim_{x\to0}\frac{\sin\left(\dfrac{x^2}2\right)}{\sqrt2\,\sin^2\left(\dfrac x2\right)} =\lim_{x\to0}\frac{\dfrac{x^2}2}{\sqrt2\,\left(\dfrac x2\right)^2}=\sqrt2.$$