Q45 / 90 · NTA Abhyas (JEE Main) 2020 · Mathematics · Limits

The value of limx→0+⁡xx2+xxx is equal to

  • A
    0
  • Correct answer
    B
    1
  • C
    2
  • D
    12

Solution

limx→0+⁡xx2=elimx→0+⁡x2ln⁡x=eL (say)
L=limx→0+⁡ln⁡x1x2=limx→0+⁡1x-2x3 {applying L’ hospital rule}
=limx→0+⁡-x22=0
Hence, limx→0+⁡xx2=e0=1

Similarly, limx→0+⁡xx=elimx→0+⁡xlnx=elimx→0+⁡lnx1x

=elimx→0+⁡1x-1x2=elimx→0+⁡-x=e0=1


Hence, limx→0+⁡xxx=0

Therefore, limx→0+⁡xx2+limx→0+⁡xxx=1

Hence, the correct answer is option: 'B'.

More Limits Questions