Question

Evaluate the Limit x -[x] / x-2 when x tends to 2 negative .​

Answer 1

$\large\underline{\sf{Given \:Question - }}$

Evaluate the following limit :-

$\rm :\longmapsto\:\displaystyle\lim_{x \to 2^ - } \frac{x - [x]}{x - 2}$

$\red{\large\underline{\sf{Solution-}}}$

Given expression is

$\rm :\longmapsto\:\displaystyle\lim_{x \to 2^ - } \frac{x - [x]}{x - 2}$

To evaluate this limit, we use method of Substitution.

So, we substitute

$\red{\rm :\longmapsto\:x = 2 - h \: \: as \: x \to \: 2 \: \: so \: h \to \: 0 \: }$

So, above expression reduced to

$\rm \: = \:\displaystyle\lim_{h \to \: 0} \frac{2 - h - [2 - h]}{2 - h - 2}$

$\rm \: = \:\displaystyle\lim_{h \to \: 0} \frac{2 - h - [2 - h]}{ - h}$

We know, by definition of Greatest Integer function,

$\red{\rm :\longmapsto\:\boxed{ \tt{ \: 1 \leqslant x < 2 \: \rm \implies\:[x] = 1 \: }}}$

So, we get

$\rm \: = \:\displaystyle\lim_{h \to \: 0} \frac{2 - h - 1}{ - h}$

$\rm \: = \:\displaystyle\lim_{h \to \: 0} \frac{1 - h}{ - h}$

$\rm \: = \:\dfrac{1 - 0}{ - 0}$

$\rm \: = \: - \: \infty$

Hence,

$\purple{\bf\implies \:\:\boxed{ \tt{ \: \displaystyle\lim_{x \to 2^ - } \frac{x - [x]}{x - 2} = - \: \infty \: }}}$

More to know :-

$\boxed{ \tt{ \: \displaystyle\lim_{x \to 0} \frac{sinx}{x} = 1 \: }}$

$\boxed{ \tt{ \: \displaystyle\lim_{x \to 0} \frac{tanx}{x} = 1 \: }}$

$\boxed{ \tt{ \: \displaystyle\lim_{x \to 0} \frac{log(1 + x)}{x} = 1 \: }}$

$\boxed{ \tt{ \: \displaystyle\lim_{x \to 0} \frac{ {e}^{x} - 1}{x} = 1 \: }}$

$\boxed{ \tt{ \: \displaystyle\lim_{x \to 0} \frac{ {a}^{x} - 1}{x} = loga \: }}$