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NTA Abhyas 2020
Exam
NTA Abhyas
Year
2020
Subject
Mathematics
Chapter
Limits

NTA Abhyas 2020 · Mathematics · Limits

The value of _ x → 1 xtan\ x\ x - 1 ~is equal to (where \ x\...

The value of lim⁡x→1xtan{x}x−1\lim_{x \to 1}\dfrac{xtan\left\{x\right\}}{x - 1}~is equal to (where \left\{x\right\}denotesthefractionalpartofdenotes the fractional part ofx$)

Options

  1. A−1- 1
  2. B00
  3. C11
  4. DDoes not exist

Correct answer

D. Does not exist

Step-by-step solution

LHL=lim⁡x→1−xtan{x}x−1LHL = \lim_{x \to {1}^{-}}\dfrac{xtan\left\{x\right\}}{x - 1}
Let, x=1−h,asx→1−,h→0+x = 1 - h, asx \to {1}^{-}, h \to {0}^{+}
⇒LHL=lim⁡h→0+(1−h)tan⁡{1−h}−h⇒ LHL = \lim_{h \to {0}^{+}}\dfrac{\left(1 - h\right)\tan \left\{1 - h\right\}}{- h}
∴LHL=lim⁡h→0+(1−h)−htan(1)=−∞∴ LHL = \lim_{h \to {0}^{+}}\left(1 - h\right)- htan\left(1\right) = - \infty
Now, RHL=lim⁡x→1+xtan{x}x−1RHL = \lim_{x \to {1}^{+}}\dfrac{xtan\left\{x\right\}}{x - 1}
Let, x=1+h,asx→1+,h→0+x = 1 + h, asx \to {1}^{+}, h \to {0}^{+}
⇒RHL=lim⁡h→0+(1+h)tan⁡{1+h}h⇒ RHL = \lim_{h \to {0}^{+}}\dfrac{\left(1 + h\right)\tan \left\{1 + h\right\}}{h}
⇒RHL=lim⁡h→0+(1+h)tanhh=lim⁡h→0+(1+h)⇒ RHL = \lim_{h \to {0}^{+}}\dfrac{ \left(1 + h\right)tanh}{h} = \lim_{h \to {0}^{+}}\left(1 + h\right)
∴RHL=(1+0)=1∴ RHL = \left(1 + 0\right) = 1
Since, LHL≠RHLLHL \ne RHL
∴∴ the limit of the function does not exist at x=1x = 1
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