JEE MATHEMATICS MCQ: LIMITS

1 Easy

1 of 1

Application +4 | -1

Let

f(x)= \begin{cases}|x-3| & \text { if } x \geq 1 \\ \frac{x^2}{4}-\frac{3 x}{2}+\frac{13}{4} & \text { if } x<1\end{cases}

Then

\lim _{x \rightarrow 1} f(x)

is equal to

View answer and explanation

Correct option: D

Option D (2) is correct because the left-hand limit and right-hand limit at x=1 both equal 2, hence the limit exists and is equal to 2.

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