If $f:R \to R$ is defined by $f\left( x \right) = \left\{ {\begin{array}{*{20}{c}} {\frac{{x - 2}}{{{x^2} - 3x + 2}}}&{,{\rm{if}}}&{x \in R - \left\{ {1,2} \right\}}\\ 2&{,{\rm{if}}}&{x = 1}\\ 1&{,\,{\rm{if}}}&{x = 2} \end{array}} \right.$ Then $\mathop {\lim }\limits_{x \to 2} \frac{{f\left( x \right) - f\left( 2 \right)}}{{x - 2}}$ is equal to

JEE MATHEMATICS MCQ: LIMITS

1 Easy

1 of 1

Application +4 | -1

f:R \to R

is defined by

f\left( x \right) = \left\{ {\begin{array}{cccccccccccccccccccc} {\frac{{x - 2}}{{{x^2} - 3x + 2}}}&{,\mathrm{if}}&{x \in R - \left\{ {1,2} \right\}}\\ 2&{,\mathrm{if}}&{x = 1}\\ 1&{,\,\mathrm{if}}&{x = 2} \end{array}} \right.

Then

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

is equal to

View answer and explanation

Correct option: B

- 1

Option B (-1) is correct because the limit of the given expression, after simplification and factorization, evaluates to -1.