Question

Evaluate the limit when x tends to 2 positive . ( ques in attachment )​

Answer 1

Answer:

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Step-by-step explanation:

Answer 2

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

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, given limit can be reduced to

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

We know,

By definition of Greatest Integer function, if

$\red{\rm :\longmapsto\:\boxed{ \tt{ \: 2 \leqslant x < 3 \: \rm \implies\:[x] = 2}}}$

So, we get

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

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

$\rm \: = \:1$

Hence,

$\purple{\bf\implies \:\:\boxed{ \tt{ \: \displaystyle\lim_{x \to 2^+} \frac{x - [x]}{x - 2} = 1 \: }}}$

More to know :-

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

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

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

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

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