2
$\begingroup$

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

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?

$\endgroup$
  • 1
    $\begingroup$ $1/x^2\to +\infty$ $\endgroup$ – A.Γ. Dec 16 '18 at 18:54
  • $\begingroup$ You can only use squeeze theorem when upper and lower bound limits exist and are equal. $\endgroup$ – D.B. Dec 16 '18 at 18:57
  • $\begingroup$ Use L'Hopital's rule. $\endgroup$ – D.B. Dec 16 '18 at 18:59
5
$\begingroup$

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 $$ \lim_{x\to0}\frac{\sin x}{x}=1 $$

$\endgroup$
2
$\begingroup$

The function is even, we assume $0<x<1$.

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

thus

$$\cos^2(x)<\frac{\sin^2(x)}{x^2}=\cos^2(c)<1$$

$\endgroup$
0
$\begingroup$

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.)

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.