Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • Engineering
  • 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 

  • Option 1)

    -1

  • Option 2)

    -2

  • Option 3)

    -3

  • Option 4)

    -4

 

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 })}

-

 

 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

View full answer

JEE Main high-scoring chapters and topics

Study 40% syllabus and score up to 100% marks in JEE

Similar Questions

Trending Articles/News

anam-khan

BTech 2025 Admission LIVE: JoSAA revised schedule out on josaa.nic.in; updates on MHT CET, KEAM counselling
anam-khan

Andhra launches free JEE, NEET coaching to over 1 lakh govt school students
anam-khan

JoSAA Counselling 2025 LIVE: BTech registration starts at josaa.nic.in for IIT, NIT, IIIT admissions

Latest Question