Lim x -6-5-2 x-3-2-x2-6

Lim_x →√(6)√(5+2x) (√(3)+√(2))/x^2 6 =lim_x →√(6)√(5+2x) (√((√(3)+√(2))^2))/x^2 6 =lim_x →√(6)√(5+2x) (√(5+2√(6)))/x^2 6 Rationalising the numenator =lim_x →√ ...

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Byju's Answer

Standard XII

Mathematics

Rationalization Method to Remove Indeterminate Form

lim x →√6√5+2...

Question

${lim}_{x \to \sqrt{6}} \frac{\sqrt{5 + 2 x} - (\sqrt{3} + \sqrt{2})}{x^{2} - 6}$

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Solution

${lim}_{x \to \sqrt{6}} \frac{\sqrt{5 + 2 x} - (\sqrt{3} + \sqrt{2})}{x^{2} - 6}$

$= {lim}_{x \to \sqrt{6}} \frac{\sqrt{5 + 2 x} - (\sqrt{(\sqrt{3} + \sqrt{2})^{2}})}{x^{2} - 6}$

$= {lim}_{x \to \sqrt{6}} \frac{\sqrt{5 + 2 x} - (\sqrt{5 + 2 \sqrt{6}})}{x^{2} - 6}$

Rationalising the numenator

$= {lim}_{x \to \sqrt{6}} \frac{\sqrt{5 + 2 x} - (\sqrt{5 + 2 \sqrt{6}})}{x^{2} - 6} \times \frac{\sqrt{5} + 2 x + (\sqrt{5 + 2 \sqrt{6}})}{\sqrt{5 + 2 x} + (\sqrt{5 + 2 \sqrt{6}})}$

$= {lim}_{x \to \sqrt{6}} \frac{5 + 2 x - 5 - 2 \sqrt{6}}{(x^{2} - 6) (\sqrt{5 + 2 x} + (\sqrt{5 + 2 \sqrt{6}}))}$

$= {lim}_{x \to \sqrt{6}} \frac{2 (x - \sqrt{6})}{(x^{2} - 6) (\sqrt{5 + 2 x} + (\sqrt{5 + 2 \sqrt{6}}))}$

$= {lim}_{x \to \sqrt{6}} \frac{2}{(x + 6) (\sqrt{5 + 2 x} + (\sqrt{5 + 2 \sqrt{6}}))}$

$= \frac{2}{(\sqrt{6} + \sqrt{6}) (\sqrt{5 + 2 \sqrt{6}} + (\sqrt{5 + 2 \sqrt{6}}))}$

$= \frac{2}{(2 \sqrt{6}) (2 \sqrt{5 + 2 \sqrt{6}})}$

$= \frac{1}{(2 \sqrt{6}) (\sqrt{5 + 2 \sqrt{6}})}$

$= \frac{1}{(2 \sqrt{6}) (\sqrt{3} + \sqrt{2})}$

$= \frac{(\sqrt{3} - \sqrt{2})}{(2 \sqrt{6})}$

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limx→√6√5+2x−(√3+√2)x2−6

${lim}_{x \to \sqrt{6}} \frac{\sqrt{5 + 2 x} - (\sqrt{3} + \sqrt{2})}{x^{2} - 6}$