Question

i need answer of this question ​

Answer 1

Answer:

Step-by-step explanation:

We have,

$\tt{f(x)=\begin{cases}\tt{\dfrac{k\,sin(x)}{x}},\,\,\,\tt{x}\neq0\\\,\,\,\,\,\,\,\,\,\tt{2},\,\,\,\,\,\,\,\,\,\,\,\,\,\,\tt{x}=0\end{cases}}$

Since it is continuous at x = 0, so,

$\sf{\displaystyle\lim_{x\to0}\,f(x)=2}$

$\sf{\implies\displaystyle\lim_{x\to0}\,\dfrac{k\,sin(x)}{x}=2}$

$\sf{\implies\,k\displaystyle\lim_{x\to0}\,\dfrac{sin(x)}{x}=2}$

$\sf{\implies\,k\times1=2}$

$\sf{\implies\,k=2}$