Question

If $$f\left(x\right)=\{\begin{array}{ll}\frac{\sin 3x}{x},when& x\ne 0\\ 1,when& x=0\end{array}$$

Find whether f(x) is continuous at x = 0.

 

Answer 1

Given 

$$f\left(x\right)=(\genfrac{}{}{0ex}{}{\frac{\sin 3x}{x},whenx\ne 0}{1,whenx=0})$$

We observe 

(LHL at x = 0) = 

$$\underset{x\to 0^{-}}{lim}f\left(x\right)=\underset{h\to 0}{lim}f(0-h)=\underset{h\to 0}{lim}f(-h)$$

$$\underset{h\to 0}{lim}\frac{\sin (-3h)}{-h}=\underset{h\to 0}{lim}\frac{-\sin \left(3h\right)}{-h}=\underset{h\to 0}{lim}\frac{3\sin \left(3h\right)}{3h}=3\underset{h\to 0}{lim}\frac{\sin \left(3h\right)}{3h}=3\cdot 1=3$$

(RHL at x = 0) = 

$$\underset{x\to 0^{+}}{lim}f\left(x\right)=\underset{h\to 0}{lim}f\left(h\right)$$

$$\underset{h\to 0}{lim}\frac{\sin 3h}{h}=\underset{h\to 0}{lim}\frac{3\sin 3h}{3h}=3\underset{h\to 0}{lim}\frac{\sin \left(3h\right)}{3h}=3\cdot 1=3$$

Given:

$$f\left(0\right)=1$$

It is known that for a function 

$$f\left(x\right)$$  to be continuous at x = a, 

$$\underset{x\to a^{-}}{lim}f\left(x\right)=\underset{x\to a^{+}}{lim}f\left(x\right)=f\left(a\right)$$

But here,

$$\underset{x\to 0^{-}}{lim}f\left(x\right)=\underset{x\to 0^{+}}{lim}f\left(x\right)\ne f\left(0\right)$$

Hence

$$f\left(x\right)$$

$$x=0$$