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If-f-x-xtan-1-1-x-x-0-0-x-0-show-that-f-is-countinous-but-not-differentiable-at-x-0-

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Question Number 11608 by agni5 last updated on 29/Mar/17
If f(x)=xtan^(−1) ((1/x)) , x≠0 =0 , x=0 show that f is countinous but not differentiable at x=0.
$$\mathrm{If}\:\mathrm{f}\left(\mathrm{x}\right)=\mathrm{xtan}^{−\mathrm{1}} \left(\frac{\mathrm{1}}{\mathrm{x}}\right)\:,\:\:\:\:\:\:\:\mathrm{x}\neq\mathrm{0} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:=\mathrm{0}\:,\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{x}=\mathrm{0} \\ $$$$\mathrm{show}\:\mathrm{that}\:\mathrm{f}\:\mathrm{is}\:\mathrm{countinous}\:\mathrm{but}\:\mathrm{not}\:\mathrm{differentiable} \\ $$$$\mathrm{at}\:\mathrm{x}=\mathrm{0}. \\ $$
Answered by mrW1 last updated on 30/Mar/17
f(0)=0 lim_(x→0) f(x)=lim_(x→0) xtan^(−1) ((1/x))=(lim_(x→0) x)×(lim_(x→0) tan^(−1) (1/x)) =0×(±(π/2))=0 since f(0)=lim_(x→0) f(x) ⇒f(x) is continous at x=0. f′(x)=tan^(−1) ((1/x))+x((1/(1+(1/x^2 ))))(−(1/x^2 ))=tan^(−1) ((1/x))−(x/(1+x^2 )) lim_(x→−0) f′(x)=−(π/2)−0=−(π/2) lim_(x→+0) f′(x)=(π/2)−0=(π/2) since lim_(x→−0) f′(x)≠lim_(x→+0) f′(x) ⇒f(x) is not differentiable at x=0.
$${f}\left(\mathrm{0}\right)=\mathrm{0} \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:{f}\left({x}\right)=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:{x}\mathrm{tan}^{−\mathrm{1}} \left(\frac{\mathrm{1}}{{x}}\right)=\left(\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:{x}\right)×\left(\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\mathrm{tan}^{−\mathrm{1}} \frac{\mathrm{1}}{{x}}\right) \\ $$$$=\mathrm{0}×\left(\pm\frac{\pi}{\mathrm{2}}\right)=\mathrm{0} \\ $$$${since}\:{f}\left(\mathrm{0}\right)=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:{f}\left({x}\right) \\ $$$$\Rightarrow{f}\left({x}\right)\:{is}\:{continous}\:{at}\:{x}=\mathrm{0}. \\ $$$$ \\ $$$${f}'\left({x}\right)=\mathrm{tan}^{−\mathrm{1}} \left(\frac{\mathrm{1}}{{x}}\right)+{x}\left(\frac{\mathrm{1}}{\mathrm{1}+\frac{\mathrm{1}}{{x}^{\mathrm{2}} }}\right)\left(−\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\right)=\mathrm{tan}^{−\mathrm{1}} \left(\frac{\mathrm{1}}{{x}}\right)−\frac{{x}}{\mathrm{1}+{x}^{\mathrm{2}} } \\ $$$$\underset{{x}\rightarrow−\mathrm{0}} {\mathrm{lim}}\:{f}'\left({x}\right)=−\frac{\pi}{\mathrm{2}}−\mathrm{0}=−\frac{\pi}{\mathrm{2}} \\ $$$$\underset{{x}\rightarrow+\mathrm{0}} {\mathrm{lim}}\:{f}'\left({x}\right)=\frac{\pi}{\mathrm{2}}−\mathrm{0}=\frac{\pi}{\mathrm{2}} \\ $$$${since}\:\underset{{x}\rightarrow−\mathrm{0}} {\mathrm{lim}}\:{f}'\left({x}\right)\neq\underset{{x}\rightarrow+\mathrm{0}} {\mathrm{lim}}\:{f}'\left({x}\right) \\ $$$$\Rightarrow{f}\left({x}\right)\:{is}\:{not}\:{differentiable}\:{at}\:{x}=\mathrm{0}. \\ $$

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