# Using the Squeeze Theorem on $\lim_{x\to 0}\frac{\sin^2x}{x^2}$

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

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?

• $1/x^2\to +\infty$ – A.Γ. Dec 16 '18 at 18:54
• You can only use squeeze theorem when upper and lower bound limits exist and are equal. – D.B. Dec 16 '18 at 18:57
• Use L'Hopital's rule. – D.B. Dec 16 '18 at 18:59

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.

By MVT, $$\sin(x)=x\cos(c)$$ with $$0.

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.)