limits - Using the Squeeze Theorem on $lim_{xto 0}frac{sin^2x}{x^2}$

$$\lim_{x\to 0}\frac{\sin^2x}{x^2}$$

I'm trying to evaluate this limit using Squeeze Theorem. However, looking at the graph I know it approaches $1$, but I am getting $0$ using the Squeeze Theorem.

$$-\frac{1}{x^2} < \frac{\sin^2x}{x^2} < \frac{1}{x^2}$$

when I sub in $0$ it's just $0$. What am I doing wrong?

Edit: Wait, it's not zero! The upper and lower bounds are indeterminate. So I can't use squeeze theorem, correct?