A Squeezed Limit

Calculus Level 2

$\large \lim_{x \to \infty} \frac{\sin^2 x}{x} = \, ?$

0 1

\infty

Limit does not exist.

3 solutions

Andrew Ellinor
Dec 15, 2015

Because $0 \leq \sin^2x \leq 1$ for all $x > 1,$ we have that $0 \leq \frac{\sin^2x}{x} \le \frac{1}{x}.$

As $x$ approaches infinity, the right hand side of this inequality tends toward 0, thereby causing $\dfrac{\sin^2x}{x}$ to also tend toward 0.

Good use of the Squeeze Theorem

Jesus Ramirez - 5 years, 5 months ago

using L,Hospitals Rule FUnction Gives INfinity/INfinity Form... so,After Differtiating WRT x Gives 2sinxcosx/1 now substitute x tending value,.. so,...Answer Is Infinity..

Vamsi Krishna - 1 year, 8 months ago

That's not true because sin^2(x) x-> just jump betweens 1 and 0 while x becomes infinitely large. So it's 0. You cant use L'Hopital unless both sides are approaching either 0 or infinity.

Bora Kargı - 1 year, 8 months ago

You cannot apply L' Hospital rule here as it is not of the form $\dfrac{\infty}{\infty}$ or $\dfrac{0}{0}$. Even if you apply L' Hospital rule, the limit would be not defined since $2\sin x\cos x\leq 2$ and limit for both $\sin x$ and $\cos x$ does not exist at $\infty$.

devansh kamra - 9 months, 3 weeks ago

Aleksa Radovanović
Jan 9, 2016

There's a theorem that states: Product of bounded sequence and zero sequence is 0.

Bostang Palaguna
Oct 12, 2020

notice that as $x$ goes to infinity, $\frac{1}{x}$ goes to $0$

since $sin^2(x)$ will be just oscillating between 0 and 1,

we conclude the limit of the given function as $x$ goes to infinity is $\boxed{0}$

A Squeezed Limit

3 solutions

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