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  • Solve it, - Limit , continuity and differentiability - JEE Main-6

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

  • Option 1)

    -2

  • Option 2)

    -1

  • Option 3)

    0

  • Option 4)

    1

 

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{e^{|-h|}+|\sin (-h)|-1}{-h}

=\lim_{h\rightarrow 0^{+}}\frac{e^{-h}-1\sin (h)}{-h}  =\lim_{h\rightarrow 0^{+}}\frac{(e^{h}-1)}{h}-\lim_{h\rightarrow 0^{+}}\frac{\sin h}{h}

=-1-1=-2

 

 

 

 

 

 


Option 1)

-2

Option 2)

-1

Option 3)

0

Option 4)

1

Posted by

Himanshu

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