Question

The value of k which makes $f\left(x\right)=\{\begin{array}{c}\sin \frac{1}{x},x\ne 0\\ k,x=0\end{array}$ continuous at $x=0$ is?

• A. $8$
• B. $1$
• C. $-1$
• D. None

Answer 1

Answer: (D)

None

Then k=?

$\begin{array}{c}\Rightarrow f\left(0\right)=k\\ \Rightarrow {f\left(0^{+}\right)}_{limx\to 0^{+}}={f\left(0^{+h}\right)}_{limh\to 0}={\sin }_{limh\to 0}\frac{1}{(o+h)}=\sin 0\\ \Rightarrow {f\left(0^{-}\right)}_{limx\to 0^{-}}={f\left(0^{-h}\right)}_{limh\to 0}={\sin }_{limh\to 0}\frac{1}{(o-h)}=\sin 0\\ \Rightarrow f\left(0\right)=f\left(0^{+}\right)=f\left(0^{-}\right)=f\left(0\right)=k=0\end{array}$

the value of $k$ is zero