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  • Stuck here, help me understand: - Limit , continuity and differentiability - JEE Main-11

Left hand derivative of  f(x) = \sin |x| + |x|  at x=0 equals 

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  • Option 2)

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  • Option 3)

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  • Option 4)

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Answers (1)

best_answer

 As we have learned

Left Hand Derivative -

Left hand derivative of  f(x) at  x = x0  is given by

f'(x_{\circ })=\lim_{h\rightarrow \circ }\:\:\frac{f(x_{\circ }-h)-f(x_{\circ })}{(x_{\circ }-h)-(x_{\circ })}Lf'(x_{\circ })=\lim_{h\rightarrow \circ }\:\:\frac{f(x_{\circ }-h)-f(x_{\circ })}{(x_{\circ }-h)-(x_{\circ })}

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 LHD=\lim_{h\rightarrow 0^{+}}\frac{f(0-h)-f(0)}{(0-h)-0}= \lim_{h\rightarrow 0^{+}}\frac{\sin |-h| + |-h |-0}{-h}

=\lim_{h\rightarrow 0^{+}}\frac{\sin |h|+|h|}{-h} = \lim_{h\rightarrow 0^{+}}\frac{\sin h + h }{h}

=\lim_{h\rightarrow 0^{+}}\frac{-\sin h}{h} - \lim_{h\rightarrow 0^{+}}\frac{h}{h}=-1-1=-2

 

 

 

 


Option 1)

-1

Option 2)

-2

Option 3)

-3

Option 4)

-4

Posted by

Himanshu

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