Q26 / 90 · NTA Abhyas (JEE Main) 2020 · Mathematics · Limits

The value of limx→1xtanxx-1 is equal to (where x denotes the fractional part of x)

  • A
    -1
  • B
    0
  • C
    1
  • Correct answer
    D
    Does not exist

Solution

LHL=limx→1-xtanxx-1
Let, x=1-h, as x→1-, h→0+
⇒LHL=limh→0+1-htan1-h-h
∴LHL=limh→0+1-h-htan1=-∞
Now, RHL=limx→1+xtanxx-1
Let, x=1+h, as x→1+, h→0+
⇒RHL=limh→0+1+htan1+hh
⇒RHL=limh→0+1+htanhh=limh→0+1+h
∴RHL=1+0=1
Since, LHL≠RHL
∴ the limit of the function does not exist at x=1

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

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