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Continuity — Maths STD 12 Science — Question

Maharashtra BoardEnglish MediumSTD 12 ScienceMathsContinuity5 Marks
Question
If $\text{f}\text{(x)}=\begin{cases}\frac{1-\cos\text{x}}{\text {x}^2}, & \text{when} \text{ x}\neq 0\\1, & \text{when}\text{ x} = 0\end{cases}$ Show that f(x) is discontinuous at x = 0.
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Answer

Given,
$\text{f}\text{(x)}=\frac{1-\cos \text{x}}{\text{x}^2}, \text{when x}\neq0$
$\text{f}\text{(x)}=1, \text{when x}=0$
consider
$\lim\limits_{\text{x} \rightarrow 0}\text{f}\text{(x)}=\lim\limits_{\text{x} \rightarrow 0}\frac{1-\cos \text{x}}{\text{x}^2}$
$\lim\limits_{\text{x} \rightarrow 0}\text{f}\text{(x)}=\lim\limits_{\text{x} \rightarrow 0}\frac{2\sin^2\frac{\text{x}}{2}}{\text{x}^2}$
$\lim\limits_{\text{x} \rightarrow 0}\text{f}\text{(x)}=\lim\limits_{\text{x} \rightarrow 0}\frac{2\sin^2\frac{\text{x}}{2}}{\frac{4\text{x}^2}{4}}$
$\lim\limits_{\text{x} \rightarrow 0}\text{f}\text{(x)}=\lim\limits_{\text{x} \rightarrow 0}\frac{2\sin\frac{\text{x}^2}{2}}{\frac{4\text{x}^2}{2}}$
$\lim\limits_{\text{x} \rightarrow 0}\text{f}\text{(x)}=\frac{2}{4}\lim\limits_{\text{x} \rightarrow 0}\frac{\sin\frac{\text{x}}{2}}{\frac{\text{x}}{2}}$
$\lim\limits_{\text{x} \rightarrow 0}\text{f}\text{(x)}=\frac{1}{2}(1)$
Given f(0) = 1
$\therefore\lim\limits_{\text{x} \rightarrow 0}\text{f}\text{(x)}\neq\text{f}(0)$
Thus, f(x) is discontinuous at x = 0.

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