Question

formula for finding multipliar​

Answer 1

Given Limit:-

$\\\bullet\quad\displaystyle \sf \lim_{x \to 2} \dfrac{x^{2} -4}{\sqrt{x+2} -\sqrt{3x-2} }$

Solution:-

By Rationalizing the denominator,

$\\\quad\longrightarrow\quad\displaystyle \sf \lim_{x \to 2} \dfrac{x^{2} -4}{\sqrt{x+2} -\sqrt{3x-2} } \times \dfrac{\sqrt{x+2} +\sqrt{3x-2}}{\sqrt{x+2} +\sqrt{3x-2}}$

$\\\quad\longrightarrow\quad\displaystyle \sf \lim_{x \to 2} \dfrac{(x+2)(x-2)[\sqrt{x+2}+\sqrt{3x-2} \:] }{(\sqrt{x+2})^{2} -(\sqrt{3x-2})^{2} }$

$\\\quad\longrightarrow\quad\displaystyle \sf \lim_{x \to 2} \dfrac{(x+2)(x-2)[\sqrt{x+2}+\sqrt{3x-2} \:] }{ x+2-3x+2 }$

$\\\quad\longrightarrow\quad\displaystyle \sf \lim_{x \to 2} \dfrac{(x+2)(x-2)[\sqrt{x+2}+\sqrt{3x-2} \:] }{ 4-2x }$

$\\\quad\longrightarrow\quad\displaystyle \sf \lim_{x \to 2} \dfrac{(x+2)(x-2)[\sqrt{x+2}+\sqrt{3x-2} \:] }{ -2(x-2) }$

$\\\quad\longrightarrow\quad\displaystyle \sf \lim_{x \to 2} \dfrac{(x+2)[\sqrt{x+2}+\sqrt{3x-2} \:] }{ -2 }$

$\\\quad\longrightarrow\quad\sf \dfrac{(2+2)[\sqrt{2+2\:}+\sqrt{3(2)-2\:} }{-2}$

$\\\quad\longrightarrow\quad\sf \dfrac{4\times4}{-2}$

$\\\quad\therefore\quad \boxed{\frak{-8} }\bigstar$

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