
If $f\left( x \right)=x\sin \dfrac{1}{x},x\ne 0$ then the value of the function at $x=0$ so that the function is the continuous at $x=0$ is,
A. 1
B. 0
C. – 1
D. Indeterminate
Answer
605.1k+ views
Hint: We know that if a function $f\left( x \right)$ is continuous at a point $x=a$, then $\underset{x\to a}{\mathop{\lim }}\,\left( f\left( x \right) \right)$ must be equal to $f\left( a \right)$. So, we will find the limit of the function at $x=0$. We will use the sandwich theorem to find the limit.
Complete step-by-step answer:
We have been given $f\left( x \right)=x\sin \dfrac{1}{x},x\ne 0$ and asked to find the value of $f\left( x \right)$. So, it is continuous at $x=0$.
We know that if a function $f\left( x \right)$ is continuous at $x=a$, then
$\begin{align}
& \underset{x\to a}{\mathop{\lim }}\,f\left( x \right)=f\left( a \right) \\
& \Rightarrow \underset{x\to 0}{\mathop{\lim }}\,x\sin \dfrac{1}{x}=f\left( 0 \right) \\
\end{align}$
Sandwich theorem is also known as squeeze theorem in which if we have three function say f, g, h and $f\left( x \right)\le g\left( x \right)\le h\left( x \right)$ then at any point $x=\alpha $. We have $f\left( \alpha \right)=h\left( \alpha \right)$ then $g\left( k \right)$ must be equal to them.
Now, we will use the sandwich theorem to find the above limit. We know that $\sin x$ oscillates between $\pm 1$, that is,
$-1\le \sin x\le 1$
On multiplying both sides by x, then depending upon the sign of x, we get,
If $x>0$
$\Rightarrow -x\le x\sin \left( \dfrac{1}{x} \right)\le x$
If $x<0$
$\Rightarrow -x\ge x\sin \left( \dfrac{1}{x} \right)\ge x$
Since,
$\begin{align}
& \underset{x\to {{0}^{+}}}{\mathop{\lim }}\,x=\underset{x\to {{0}^{+}}}{\mathop{\lim }}\,\left( -x \right)=0 \\
& \underset{x\to {{0}^{-}}}{\mathop{\lim }}\,x=\underset{x\to {{0}^{-}}}{\mathop{\lim }}\,-x=0 \\
\end{align}$
According, to sandwich theorem we can say,
$\begin{align}
& \underset{x\to 0}{\mathop{\lim }}\,x\sin \left( \dfrac{1}{x} \right)=0 \\
& \Rightarrow f\left( 0 \right)=0 \\
\end{align}$
Therefore, the value of the function at $x=0$ is equal 0 and the correct option is option B.
Note: Be careful while finding the limit and take care of the sign during the inequality. Also, remember that we have used the sandwich theorem to solve this question, which is the best way to approach this type of questions in exams. It makes it easier to solve such difficult limits in less time. So, students must be familiar with this theorem as it will help in solving fast and scoring marks too.
Complete step-by-step answer:
We have been given $f\left( x \right)=x\sin \dfrac{1}{x},x\ne 0$ and asked to find the value of $f\left( x \right)$. So, it is continuous at $x=0$.
We know that if a function $f\left( x \right)$ is continuous at $x=a$, then
$\begin{align}
& \underset{x\to a}{\mathop{\lim }}\,f\left( x \right)=f\left( a \right) \\
& \Rightarrow \underset{x\to 0}{\mathop{\lim }}\,x\sin \dfrac{1}{x}=f\left( 0 \right) \\
\end{align}$
Sandwich theorem is also known as squeeze theorem in which if we have three function say f, g, h and $f\left( x \right)\le g\left( x \right)\le h\left( x \right)$ then at any point $x=\alpha $. We have $f\left( \alpha \right)=h\left( \alpha \right)$ then $g\left( k \right)$ must be equal to them.
Now, we will use the sandwich theorem to find the above limit. We know that $\sin x$ oscillates between $\pm 1$, that is,
$-1\le \sin x\le 1$
On multiplying both sides by x, then depending upon the sign of x, we get,
If $x>0$
$\Rightarrow -x\le x\sin \left( \dfrac{1}{x} \right)\le x$
If $x<0$
$\Rightarrow -x\ge x\sin \left( \dfrac{1}{x} \right)\ge x$
Since,
$\begin{align}
& \underset{x\to {{0}^{+}}}{\mathop{\lim }}\,x=\underset{x\to {{0}^{+}}}{\mathop{\lim }}\,\left( -x \right)=0 \\
& \underset{x\to {{0}^{-}}}{\mathop{\lim }}\,x=\underset{x\to {{0}^{-}}}{\mathop{\lim }}\,-x=0 \\
\end{align}$
According, to sandwich theorem we can say,
$\begin{align}
& \underset{x\to 0}{\mathop{\lim }}\,x\sin \left( \dfrac{1}{x} \right)=0 \\
& \Rightarrow f\left( 0 \right)=0 \\
\end{align}$
Therefore, the value of the function at $x=0$ is equal 0 and the correct option is option B.
Note: Be careful while finding the limit and take care of the sign during the inequality. Also, remember that we have used the sandwich theorem to solve this question, which is the best way to approach this type of questions in exams. It makes it easier to solve such difficult limits in less time. So, students must be familiar with this theorem as it will help in solving fast and scoring marks too.
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