$$\lim_{x\to 0}\frac{\sin^2x}{x^2}$$
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?
The lower and upper bounds you write are right, but unfortunately the lower bound has limit $-\infty$ and the upper bound has limit $\infty$, so they can't be used to determine the given limit.
If you want to apply squeezing, you can prove geometrically that $$ \cos^2x<\frac{\sin^2x}{x^2}<\frac{1}{\cos^2x} $$ which is basically the usual proof that $$ \lim_{x\to0}\frac{\sin x}{x}=1 $$
The function is even, we assume $0<x<1$.
By MVT, $$\sin(x)=x\cos(c)$$ with $0<c<x$.
thus
$$\cos^2(x)<\frac{\sin^2(x)}{x^2}=\cos^2(c)<1$$
As you found out, the upper and lower bounds are indeterminate. Therefore, you cannot determine the limit that way. Instead, you can refer to the following well-known limit
$$\lim_{x \to 0}\frac{\sin x}{x} = 1$$
If you want to use the Squeeze Theorem, you can refer to the geometric proof of the limit above. (Then, you can apply the geometric proof here and reach a conclusion by the Squeeze Theorem.)
