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:
L = \(\rm\lim_{x\to 1}{\ln x +x-1\over x^2 -1}\)
On putting the limits L = \(0\over 0\)
Applying L'Hospital Rule
L = \(\rm\lim_{x\to 1}{{1\over x} +1\over 2x}\)
Putting the limits we get,
L = \(\rm{{1\over 1} +1\over 2(1)}\)
L = 1
