Question

How do you find the limit of $\mathop {\lim }\limits_{x \to 0} {\text{ }}\dfrac{{{{\sin }^2}x}}{x}$ ?

Answer 1

Hint : Here we are asked to find the limit of $\dfrac{{{{\sin }^2}x}}{{\sin x}}$ as $x \to 0$ so we will solve this by using the following limit identity $\mathop {\lim }\limits_{x \to 0} {\text{ }}\dfrac{{\sin x}}{x} = 1$ . So we must know that property of a sine function while calculating the above problem that it lies between $- 1$ and $1$ . Lastly after multiplying we will transform the given function into one of standard limit properties and then by applying the squeeze theorem we can find the limit of the function.
Formula: We will solve the above question by using Squeeze theorem and with the help of following limit identity
$\mathop {\lim }\limits_{x \to 0} {\text{ }}\dfrac{{\sin x}}{x} = 1$

Complete step by step solution:
As we know the following limit identity $\mathop {\lim }\limits_{x \to 0} {\text{ }}\dfrac{{\sin x}}{x} = 1$
Here the given function can be re written for making the use of the given fact
$\mathop {\lim }\limits_{x \to 0} {\text{ }}\dfrac{{\sin x}}{x} = 1$
So we can rewrite the question as $\mathop {\lim }\limits_{x \to 0} {\text{ }}\dfrac{{{{\sin }^2}x}}{x}$
Then we can see that $\dfrac{{\sin x}}{x}$ can be isolated from the given question
$\mathop {\lim }\limits_{x \to 0} {\text{ }}\dfrac{{\sin x}}{x}(\sin x)$
Later on we can multiply the limits a followed
$\mathop {\lim }\limits_{x \to 0} {\text{ }}\dfrac{{\sin x}}{x} \times \mathop {\lim }\limits_{x \to 0} \sin x$
As first part just equals to $1$ so after simplification we get
$\mathop {\lim }\limits_{x \to 0} {\text{ }}\sin x$
Lastly by plugging in $0$ for $x$ we can evaluate the limit
$\sin (0) = 0$
Therefore the function should approach $0$ at $x = 0$ which means $0$ is our required limit.
So, the correct answer is “0”.

Note : As we have found the limits of $x$ approaching $0$ from the positive side so there can be a doubt that what will be the limit of $x$ approaching $0$ from the negative side. So keep in mind that Squeeze theorem works on both the sides therefore $\mathop {\lim }\limits_{x \to 0} {\text{ }}\dfrac{{\sin x}}{x} = 1$ . Remember while evaluating a limit we look at a function as it approaches a specific point. The function itself gets closer and closer to a particular value when we approach a particular value of $x$ .