Question

$\sf\fbox\pink{❥︎QuestioN}$
$\sf\red{lim} \: x→2 \: \red{ \frac{( {x}^{2} - 4) }{ \sqrt{3}x - 2 - \sqrt{x} + 2}}$


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Answer 1

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

Given expression is

$\rm :\longmapsto\:\displaystyle\lim_{x \to 2}\rm \frac{ {x}^{2} - 4}{ \sqrt{3x - 2} - \sqrt{x + 2} }$

If we substitute directly x = 2, we get

$\rm \: = \: \dfrac{ {2}^{2} - 4}{ \sqrt{3(2) - 2} - \sqrt{2 + 2} }$

$\rm \: = \: \dfrac{ 4 - 4}{ \sqrt{6 - 2} - \sqrt{4} }$

$\rm \: = \: \dfrac{0}{ \sqrt{4} - 2}$

$\rm \: = \: \dfrac{0}{ 2 - 2}$

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

which is indeterminant form.

So,

$\rm :\longmapsto\:\displaystyle\lim_{x \to 2}\rm \frac{ {x}^{2} - 4}{ \sqrt{3x - 2} - \sqrt{x + 2} }$

On rationalizing the denominator, we get

$\rm \: = \displaystyle\lim_{x \to 2}\rm \frac{ {x}^{2} - {2}^{2} }{ \sqrt{3x - 2} - \sqrt{x + 2} } \times \dfrac{ \sqrt{3x - 2} + \sqrt{x + 2} }{ \sqrt{3x - 2} + \sqrt{x + 2} }$

We know

$\boxed{\tt{ \: (x + y)(x - y) \: = \: {x}^{2} - {y}^{2} \: }}$

So, above can be rewritten as

$\rm \: = \: \displaystyle\lim_{x \to 2}\rm \frac{(x - 2)(x + 2)[ \sqrt{3x - 2} + \sqrt{x + 2}]}{(3x - 2) - (x + 2)}$

$\rm \: = \: \displaystyle\lim_{x \to 2}\rm \frac{(x - 2)(x + 2)[ \sqrt{3x - 2} + \sqrt{x + 2}]}{3x - 2 - x - 2}$

$\rm \: = \: \displaystyle\lim_{x \to 2}\rm \frac{(x - 2)(x + 2)[ \sqrt{3x - 2} + \sqrt{x + 2}]}{2x - 4}$

$\rm \: = \: \displaystyle\lim_{x \to 2}\rm \frac{(x - 2)(x + 2)[ \sqrt{3x - 2} + \sqrt{x + 2}]}{2(x - 2)}$

$\rm \: = \: \displaystyle\lim_{x \to 2}\rm \frac{(x + 2)[ \sqrt{3x - 2} + \sqrt{x + 2}]}{2}$

$\rm \: = \: \dfrac{(2 + 2)[ \sqrt{3(2) - 2} + \sqrt{2 + 2}]}{2}$

$\rm \: = \: \dfrac{(4)[ \sqrt{6 - 2} + \sqrt{4}]}{2}$

$\rm \: = \: 2( \sqrt{4} + 2)$

$\rm \: = \: 2(2 + 2)$

$\rm \: = \: 8$

Hence,

$\sf\implies \: \boxed{\tt{ \:\displaystyle\lim_{x \to 2}\rm \frac{ {x}^{2} - 4}{ \sqrt{3x - 2} - \sqrt{x + 2} } = 8 \: }}$

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Answer 2

Answer:

your solution is in above pics

pls mark it as brainliest