Question

Let \[f\left( x \right) = \begin{cases}\frac{1 - \cos x}{x^2}, when & x \neq 0 \\ 1 , when & x = 0\end{cases}\] Show that f(x) is discontinuous at x = 0.

 

 

Answer 1

Given: 

\[f\left( x \right) = \binom{\frac{1 - \ cosx}{x^2}, when x \neq 0}{1, when x = 0}\]

Consider:

\[\lim_{x \to 0} f\left( x \right) = \lim_{x \to 0} \left( \frac{1 - \ cosx}{x^2} \right)\]

\[ \Rightarrow \lim_{x \to 0} f\left( x \right) = \lim_{x \to 0} \left( \frac{2 \sin^2 \frac{x}{2}}{x^2} \right)\]

\[ \Rightarrow \lim_{x \to 0} f\left( x \right) = \lim_{x \to 0} \left( \frac{2 \sin^2 \frac{x}{2}}{4\left( \frac{x^2}{4} \right)} \right)\]

\[ \Rightarrow \lim_{x \to 0} f\left( x \right) = \lim_{x \to 0} \left( \frac{2 \left( \sin\frac{x}{2} \right)^2}{4 \left( \frac{x}{2} \right)^2} \right)\]

\[ \Rightarrow \lim_{x \to 0} f\left( x \right) = \frac{2}{4} \lim_{x \to 0} \left( \frac{sin\frac{x}{2}}{\frac{x}{2}} \right)^2 \]

\[ \Rightarrow \lim_{x \to 0} f\left( x \right) = \frac{1}{2} \cdot 1^2 = \frac{1}{2}\]

Given:

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

\[\lim_{x \to 0} f\left( x \right) \neq f\left( 0 \right)\]

Thus, f(x) is discontinuous at x = 0.