Question

What is the value of $\displaystyle \lim_{x \to \infty }\left( \dfrac{{{\sin }^{2}}x}{{{x}^{2}}} \right)$?

Answer 1

We need to find the value of $\displaystyle \lim_{x \to \infty }\left( \dfrac{{{\sin }^{2}}x}{{{x}^{2}}} \right)$. We find the range of the functions for numerator and denominator separately. We have $\forall x\in \mathbb{R}$, $0\le {{\sin }^{2}}x\le 1$. Therefore, we get
$0\le \dfrac{{{\sin }^{2}}x}{{{x}^{2}}}\le \dfrac{1}{{{x}^{2}}}$. We use the limit of the function to find the limit value of the function.

Complete step by step solution:
We have to find the value of $\displaystyle \lim_{x \to \infty }\left( \dfrac{{{\sin }^{2}}x}{{{x}^{2}}} \right)$.
We have to find the range of the functions individually.
We first take the trigonometric function in the numerator.
$\forall x\in \mathbb{R}$, the range of the function $f\left( x \right)=\sin x$ is $-1\le \sin x\le 1$.
Now we find the range of $g\left( x \right)={{\sin }^{2}}x$.
$\forall x\in \mathbb{R}$, the range of the function $g\left( x \right)={{\sin }^{2}}x$ is $0\le {{\sin }^{2}}x\le 1$.
We divide the both sides of the inequality $0\le {{\sin }^{2}}x\le 1$ with ${{x}^{2}}$.
The function $\left( \dfrac{{{\sin }^{2}}x}{{{x}^{2}}} \right)$ will be defined as the value of $x \to \infty$.
So, we get $\dfrac{0}{{{x}^{2}}}\le \dfrac{{{\sin }^{2}}x}{{{x}^{2}}}\le \dfrac{1}{{{x}^{2}}}\Rightarrow 0\le \dfrac{{{\sin }^{2}}x}{{{x}^{2}}}\le \dfrac{1}{{{x}^{2}}}$.
To find the limit value of $\displaystyle \lim_{x \to \infty }\left( \dfrac{{{\sin }^{2}}x}{{{x}^{2}}} \right)$, we need to find the limit value of the function
$\displaystyle \lim_{x \to \infty }\left( \dfrac{1}{{{x}^{2}}} \right)$.
We put the value of the variables and get $\displaystyle \lim_{x \to \infty }\left( \dfrac{1}{{{x}^{2}}} \right)=\dfrac{1}{\infty }=0$.
Therefore, we get as $x \to \infty$, the function $\dfrac{{{\sin }^{2}}x}{{{x}^{2}}}\to 0$.
Therefore, the value of $\displaystyle \lim_{x \to \infty }\left( \dfrac{{{\sin }^{2}}x}{{{x}^{2}}} \right)$ is 0.

Note: We need to remember that we can’t use the limit formula of $\displaystyle \lim_{x \to 0}\left( \dfrac{\sin x}{x} \right)=1$. That limit is for the limit of $x \to 0$. Here, the limit tends to infinity. Also, the value of the numerator being close to 0 doesn’t change the final value as $x \to \infty$.