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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.
Solution
Given:
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:
Thus, f(x) is discontinuous at x = 0.
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