If $f\left( 1 \right) = 2$ and $f'\left( 1 \right) = 1$, then value of $\mathop {\lim }\limits_{x \to 1} \,\,\frac{{2x - f\left( x \right)}}{{x - 1}}$ is

JEE MATHEMATICS MCQ: LIMITS

1 Easy

1 of 1

Application +4 | -1

f\left( 1 \right) = 2

and

f'\left( 1 \right) = 1

, then value of

\lim\limits_{x \to 1} \,\,\frac{{2x - f\left( x \right)}}{{x - 1}}

View answer and explanation

Correct option: C

Option C (1) is correct because by applying L'Hopital's rule and using the given values of f(1) and f'(1), the limit evaluates to 1.

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