Lim_x → 3√(x+3) √(6)/x^2 9 Rationalising the numenator =lim_x → 3(√(x+3) √(6))/(x 3)(x+3)(√(x+3) √(6))/(√(x+3)+√(6)) =lim_x → 3((x+3) 6)/(x 3)(x+3)(√(x+3)+√(6)) ...

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Byju's Answer

Standard XII

Mathematics

Rationalization Method to Remove Indeterminate Form

lim x → 3√x+3...

Question

${lim}_{x \to 3} \frac{\sqrt{x + 3} - \sqrt{6}}{x^{2} - 9}$

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Solution

${lim}_{x \to 3} \frac{\sqrt{x + 3} - \sqrt{6}}{x^{2} - 9}$

Rationalising the numenator

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

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

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

$= {lim}_{x \to 3} \frac{1}{(x + 3) (\sqrt{x + 3} + \sqrt{6})}$

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

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

$= \frac{1}{12 \sqrt{6}}$

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Similar questions

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is equal to

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limx→3√x+3−√6x2−9

limx→3√x+3−√6x2−9 Rationalising the numenator =limx→3(√x+3−√6)(x−3)(x+3)(√x+3−√6)(√x+3+√6) =limx→3((x+3)−6)(x−3)(x+3)(√x+3+√6) =limx→3(x−3)(x−3)(x+3)(√x+3+√6) =limx→31(x+3)(√x+3+√6) =1(3+3)√3+3+√6 =16(√6+√6)=16×2√6 =112√6

${lim}_{x \to 3} \frac{\sqrt{x + 3} - \sqrt{6}}{x^{2} - 9}$