Question

How do you find the limit of ${{x}^{2}}\sin \left( \dfrac{1}{x} \right)$ as x approaches 0 ?

Answer 1

To calculate the limit of the given expression, one must know that property of a sine function that it lies between -1 and 1. Using this inequality and the sandwich theorem, we can find the required limit.

Complete step by step answer:
The given expression in the question is ${{x}^{2}}\sin \left( \dfrac{1}{x} \right)$ and it is said that we have to find the limit of this expression when x approaches 0. Limit of an expression or a function at a certain point is the value of the function that it tends to take when the independent (variable x) approaches that value. To find the limit of the given expression we shall use the property of a sine function.

For any sine function $y=\sin u$, the value of the function lie between -1 and 1.
i.e. $-1\le \sin u\le 1$.
In this question, $u=\dfrac{1}{x}$
This means that $-1\le \sin \left( \dfrac{1}{x} \right)\le 1$
Now, let us multiply the above inequality by ${{x}^{2}}$.
With this we get that $-{{x}^{2}}\le {{x}^{2}}\sin \left( \dfrac{1}{x} \right)\le {{x}^{2}}$
Now, let us apply the sandwich theorem on the above inequality as the value of x approaches 0.

With this we get $\underset{x\to {{0}^{-}}}{\mathop{\lim }}\,\left( -{{x}^{2}} \right)\le \underset{x\to 0}{\mathop{\lim }}\,{{x}^{2}}\sin \left( \dfrac{1}{x} \right)\le \underset{x\to {{0}^{+}}}{\mathop{\lim }}\,\left( {{x}^{2}} \right)$ …. (i)
We know that the limit of $-{{x}^{2}}$ as the x approaches 0 from the negative x-axis is equal to 0.
$\underset{x\to {{0}^{-}}}{\mathop{\lim }}\,\left( -{{x}^{2}} \right)=0$
similarly, $\underset{x\to {{0}^{+}}}{\mathop{\lim }}\,\left( {{x}^{2}} \right)=0$
By substituting these limits in inequality (i), we get
$0\le \underset{x\to 0}{\mathop{\lim }}\,{{x}^{2}}\sin \left( \dfrac{1}{x} \right)\le 0$
This means that the value of $\underset{x\to 0}{\mathop{\lim }}\,{{x}^{2}}\sin \left( \dfrac{1}{x} \right)$ must be zero.

Therefore, the limit of ${{x}^{2}}\sin \left( \dfrac{1}{x} \right)$ as x approaches 0 is 0.

Note: Whenever you try to find the limit of a function when the value of x approaches 0, find both the limit, when x approaches to 0 from negative x-axis and when x approaches to 0 from positive x-axis. It may sometimes happen that the limit from the positive side may not be equal to the limit from the negative side, which means that limit of the function at that point does not exist.