If $f\left( x \right) = \frac{{\sin \left( {{e^{x - 2}} - 1} \right)}}{{\log \left( {x - 1} \right)}}$, then $\mathop {\lim }\limits_{x \to 2} f\left( x \right)$ is given by

JEE MATHEMATICS MCQ: LIMITS

1 Easy

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Analyzing +4 | -1

f\left( x \right) = \frac{{\sin \left( {{e^{x - 2}} - 1} \right)}}{{\log \left( {x - 1} \right)}}

, then

\lim\limits_{x \to 2} f\left( x \right)

is given by

View answer and explanation

Correct option: D

Option D (1) is correct. By using algebraic manipulation and standard limits, the limit can be evaluated to 1.

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