Search any question & find its solution
Question:
Answered & Verified by Expert
If $f(x)=\left\{\begin{array}{r}
\frac{\sin 2 x}{5 x}, \text { when } x \neq 0 \\
k, \text { when } x=0\end{array}\right.$ is continuous at $x=0$, then the value of $k$ will be
Options:
\frac{\sin 2 x}{5 x}, \text { when } x \neq 0 \\
k, \text { when } x=0\end{array}\right.$ is continuous at $x=0$, then the value of $k$ will be
Solution:
1763 Upvotes
Verified Answer
The correct answer is:
$\frac{2}{5}$
$\lim _{x \rightarrow 0} f(x)=\lim _{x \rightarrow 0} \frac{2 \sin 2 x}{2 x .5}=\frac{2}{5}=k$.
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.