# Evaluate Lim X → 2 F ( X ) (If It Exists), Where F ( X ) = ⎧ ⎪ ⎨ ⎪ ⎩ X − [ X ] , X < 2 4 , X = 2 3 X − 5 , X > 2 . - Mathematics

Evaluate $\lim_{x \to 2} f\left( x \right)$ (if it exists), where $f\left( x \right) = \left\{ \begin{array}{l}x - \left[ x \right], & x < 2 \\ 4, & x = 2 \\ 3x - 5, & x > 2\end{array} . \right.$

#### Solution

$f\left( x \right) = \begin{cases}x - \left[ x \right], & x < 2 \\ 4, & x = 2 \\ 3x - 5, & x > 2\end{cases}$
$\text{ LHL }:$
$\lim_{x \to 2^-} f\left( x \right)$
$= \lim_{x \to 2^-} \left\{ x - \left[ x \right] \right\}$
$\text{ Let } x = 2 - \text{ h, where h } \to 0 .$
$\Rightarrow \lim_{h \to 0} \left[ \left( 2 - h \right) - \left[ 2 - h \right] \right]$
$= 2 - \left( 1 \right)$
$= 1$
$\text{ RHL }:$
$\lim_{x \to 2^+} f\left( x \right)$
$= \lim_{x \to 2^+} \left( 3x - 5 \right)$
$\text{ Let } x = 2 + \text{ h, where h } \to 0 .$
$\Rightarrow \lim_{h \to 0} \left[ 3\left( 2 + h \right) - 5 \right]$
$= 6 - 5$
$= 1$
$\text{ Here }, LHL = RHL = 1$
$\therefore \lim_{x \to 2} f\left( x \right) = 1$

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#### APPEARS IN

RD Sharma Class 11 Mathematics Textbook
Chapter 29 Limits
Exercise 29.1 | Q 20 | Page 12