Evaluating $\lim_{x\to 0}{\frac{\sin^2x}{2x^2}}$ without L'Hospital

Question

I have been trying to evaluate $$\lim_{x\to 0}{\frac{\sin^2x}{2x^2}}$$ Finally, I used the L'Hospital's Theorem and I got the answer $1/2$, but I wonder if there is a way to solve this without this.
I also tried using Squeeze Theorem, but my boundaries were approaching different numbers.

Answer 1

Hint: $\sin^\prime = \cos$, so that $\sin^\prime(0) = \cos 0 = 1.$ Can you make this quantity appear?

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You can write$$\frac{\sin^2 x}{x^2} = \left(\frac{\sin x}{x}\right)^2 = \left(\frac{\sin x-\sin 0}{x-0}\right)^2 \xrightarrow[x\to 0]{} \left(\sin^\prime(0)\right)^2 = 1$$using continuity of the function $t\mapsto t^2$.

Answer 2

Hint: If you want to use the Squeeze Theorem, you can proceed geometrically and observe that for $x\neq 0$ \begin{align}\frac12|\sin x| \le \frac12|x|\le \frac12|\tan x| &\implies 1\le \frac{|x|}{|\sin x|}\le \frac{1}{|\cos x|}\\ &\implies |\cos x|\le \frac{|\sin x|}{|x|}\le 1\\&\implies\cos^2x\le \frac{\sin^2 x}{x^2}\le1\end{align}