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What is the value of \(\rm \displaystyle\lim_{x\rightarrow 1}{\log x\over x^2-1}\)
5. None of these

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Correct Answer - Option 4 : 1/2

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 = \(\rm\displaystyle\lim_{x\rightarrow 1}{\log x\over x^2-1}\)

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

Differentiating numerator and denominator wrt x

L = \(\rm \displaystyle\lim_{x\rightarrow 1}{{1\over x}\over 2x -0} = \rm\lim_{x\rightarrow 1}{1\over 2x^2}\)

Putting the value of x = 1

L = 1/2

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