Question

How do you find the limit of \[\dfrac{{{{\sin }^2}x}}{x}\] as \[x\] approaches \[0\]?

Answer 1

Hint: In the given question, we have been given to find the value of the limit of a function of sine squared divided by its argument as its argument approaches zero. We are going to use the value of the limit of sine divided by its argument as it approaches zero (which is a standard result) and then use the multiplicative property of limits for determining the value using the remaining part. Then we are going to put the value into the limit, simplify the value of the expansion, and then plug in the value of the argument and find our answer.
We are going to use the multiplicative property of limits, which is,
\[{\lim _{x \to n}}a \times b = {\lim _{x \to n}}a \times {\lim _{x \to n}}b\]

Complete step by step answer:
We have to calculate the value of \[\dfrac{{{{\sin }^2}x}}{x}\] as \[x \to 0\].
We are going to use the standard result of value of \[\dfrac{{\sin x}}{x}\] as \[x \to 0\], which is equal to \[1\].
So,
\[{\lim _{x \to 0}}\dfrac{{\sin x}}{x} = 1\]
Now, \[{\lim _{x \to 0}}\dfrac{{{{\sin }^2}x}}{x} = {\lim _{x \to 0}}\dfrac{{\sin x}}{x} \times {\lim _{x \to 0}}\sin x = 1 \times {\lim _{x \to 0}}\sin x\]
Now, putting in the value of \[x\] in \[{\lim _{x \to 0}}\sin x\], we have,
\[{\lim _{x \to 0}}\dfrac{{{{\sin }^2}x}}{x} = 1 \times 0 = 0\]

Note: In the given question, we had to calculate the value of \[\dfrac{{{{\sin }^2}x}}{x}\] as \[x \to 0\]. We solved it by using the standard relation values. Some students make mistakes when they do not remember the standard result values and get confused with it. So, it is important to know these values and how to use these values for the given question.