Search any question & find its solution
Question:
Answered & Verified by Expert
If $f(x)=\left\{\begin{array}{cl}x \sin \frac{1}{x} & , x \neq 0 \\ k & , x=0\end{array}\right.$ is continuous at
$x=0$, then the value of $k$ is
Options:
$x=0$, then the value of $k$ is
Solution:
1637 Upvotes
Verified Answer
The correct answer is:
0
If function $f(x)$ is continuous at $x=0$, then
$$
f(0)=\lim _{x \rightarrow 0} f(x)
$$
Given, $\quad f(0)=k$
$$
\begin{array}{ll}
\therefore \quad f(0)=k=\lim _{x \rightarrow 0} x \sin \frac{1}{x} \\
\Rightarrow \quad k=0 \quad\left(\because-1 \leq \sin \frac{1}{x} \leq 1\right)
\end{array}
$$
$$
f(0)=\lim _{x \rightarrow 0} f(x)
$$
Given, $\quad f(0)=k$
$$
\begin{array}{ll}
\therefore \quad f(0)=k=\lim _{x \rightarrow 0} x \sin \frac{1}{x} \\
\Rightarrow \quad k=0 \quad\left(\because-1 \leq \sin \frac{1}{x} \leq 1\right)
\end{array}
$$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.