Show that the function f(x) x^m (1 x ), &x 0 0 ,& x=0 Neirher continuous but not diffierentiable, if m 0

LHL = _ x 0^ - f(x) = _ h 0 f(0-h) = _ h 0 (-h)^m (-1 h ) = Not defined as m 0 RHL = _ x 0^ + f(x) = _ h 0 f(0+h) = _ h 0 (+h)^m (1 0+h ) =Not defined, as m 0 Since RHL and LHL are not difined, so f(x) is not continuous Let x = 0 for m 0. (LHL at x = 0) = _ x 0^ -1 f(x)-f(0) x-0 = _ h 0 f(0-h)-(0) 0-h-0 = _ h 0 (-h)^m (-1 h ) -h = _ h 0 (-h)^m-1 (-1 h ) = Not definded, as m 0 (RHL at x = 0) = _ x 0^ + f(x)-f(0) x-0 = _ h 0 f(0+h)-f(0) 0+h-0 = _ h 0 h^m (1 h ) h = _ h 0 (h)^ m^ -1 (1 h ) = Not defined m 0 Thus, f(x) is neither continuous not differentiable at x = 0 for m 0.

Show that the function f(x) x^m (1 x ), &x 0 0 ,& x=0 Neirher con…

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Question

Show that the function $\text{f(x)}\begin{cases}\text{x}^\text{m}\sin(\frac{1}{\text{x}}), &\text{x}\neq0 \\0 ,& \text{x}=0\end{cases}$
Neirher continuous but not diffierentiable, if $\text{m}\leq0$

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Answer

$\text{LHL }=\lim_\limits{\text{x}\rightarrow0^{-}}\text{f(x)}$

$=\lim_\limits{\text{h}\rightarrow0}\text{f}(0-\text{h})$

$=\lim_\limits{\text{h}\rightarrow0}(-\text{h})^\text{m}\sin\Big(-\frac{1}{\text{h}}\Big)$

= Not defined as $\text{m}\leq0$

$\text{RHL }=\lim_\limits{\text{x}\rightarrow0^{+}}\text{f(x)}$

$=\lim_\limits{\text{h}\rightarrow0}\text{f}(0+\text{h})$

$=\lim_\limits{\text{h}\rightarrow0}(+\text{h})^\text{m}\sin\Big(\frac{1}{0+\text{h}}\Big)$

=Not defined, as $\text{m}\leq0$

Since RHL and LHL are not difined, so f(x) is not continuous

Let x = 0 for $\text{m}\leq0.$

(LHL at x = 0) $=\lim_\limits{\text{x}\rightarrow0^{-1}}\frac{\text{f(x)}-\text{f}(0)}{\text{x}-0}$

$=\lim_\limits{\text{h}\rightarrow0}\frac{\text{f}(0-\text{h})-(0)}{0-\text{h}-0}$

$=\lim_\limits{\text{h}\rightarrow0}\frac{(-\text{h})^\text{m}\sin\Big(-\frac{1}{\text{h}}\Big)}{-\text{h}}$

$=\lim_\limits{\text{h}\rightarrow0}(-\text{h})^\text{m-1}\sin\Big(-\frac{1}{\text{h}}\Big)$

= Not definded, as $\text{m}\leq0$

(RHL at x = 0) $=\lim_\limits{\text{x}\rightarrow0^{+}}\frac{\text{f(x)}-\text{f}(0)}{\text{x}-0}$

$=\lim_\limits{\text{h}\rightarrow0}\frac{\text{f}(0+\text{h})-\text{f}(0)}{0+\text{h}-0}$

$=\lim_\limits{\text{h}\rightarrow0}\frac{\text{h}^\text{m}\sin\Big(\frac{1}{\text{h}}\Big)}{\text{h}}$

$=\lim_\limits{\text{h}\rightarrow0}(\text{h})^{\text{m}^{-1}}\sin\Big(\frac{1}{\text{h}}\Big)$

= Not defined $\text{m}\leq0$

Thus,

f(x) is neither continuous not differentiable at x = 0 for $\text{m}\leq0.$

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