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in Continuity and Differentiability by (114k points)
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What is the value of \(\rm\lim_{x\rightarrow 1}{\log x\over x-1}\)
1. 1
2. -1
3. 2
4. -2

1 Answer

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Best answer
Correct Answer - Option 1 : 1

Concept:

L'Hospital's Rule:

​If the limit becomes \({0\over 0}\) or \({\pm\infty\over \pm\infty}\), it is solved by differentiating numerator and denominator.

Calculation:

Let L = \(\rm\lim_{x\rightarrow 1}{\log x\over x-1}\)

Putting the value of the x, we get \({0\over 0}\).

Differentiating numerator and denominator wrt x

L = \(\rm\lim_{x\rightarrow 1}{{1\over x}\over 1}\)

Putting the value of x = 1

L = 1

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