Question

If \[f\left( x \right) = \begin{cases}\frac{\sin 3x}{x}, when & x \neq 0 \\ 1 , when & x = 0\end{cases}\]

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

 

Answer 1

Given 

\[f\left( x \right) = \binom{\frac{\sin3x}{x}, when x \neq 0}{1, when x = 0}\]

We observe 

(LHL at x = 0) = 

\[\lim_{x \to 0^-} f\left( x \right) = \lim_{h \to 0} f\left( 0 - h \right) = \lim_{h \to 0} f\left( - h \right)\]

\[\lim_{h \to 0} \frac{\sin\left( - 3h \right)}{- h} = \lim_{h \to 0} \frac{- \sin\left( 3h \right)}{- h} = \lim_{h \to 0} \frac{3\sin\left( 3h \right)}{3h} = 3 \lim_{h \to 0} \frac{\sin\left( 3h \right)}{3h} = 3 \cdot 1 = 3\]

(RHL at x = 0) = 

\[\lim_{x \to 0^+} f\left( x \right) = \lim_{h \to 0} f\left( h \right)\]

\[\lim_{h \to 0} \frac{\sin3h}{h} = \lim_{h \to 0} \frac{3\sin3h}{3h} = 3 \lim_{h \to 0} \frac{\sin\left( 3h \right)}{3h} = 3 \cdot 1 = 3\]

Given:

\[f\left( 0 \right) = 1\]

It is known that for a function 

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

\[\lim_{x \to a^-} f\left( x \right) = \lim_{x \to a^+} f\left( x \right) = f\left( a \right)\]

But here,

\[\lim_{x \to 0^-} f\left( x \right) = \lim_{x \to 0^+} f\left( x \right) \neq f\left( 0 \right)\]

Hence

\[f\left( x \right)\]

\[x = 0\]