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Limits — Maths STD 11 Science — Question

Maharashtra BoardEnglish MediumSTD 11 ScienceMathsLimits5 Marks
Question
Evaluate the following limit:
$\lim\limits_{\text{x}\rightarrow{\sqrt{6}}}\frac{\sqrt{5+2\text{x}}-\big(\sqrt{3}+\sqrt{2}\big)}{\text{x}^2-6}$
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Answer

$\lim\limits_{\text{x}\rightarrow{\sqrt{6}}}\frac{\sqrt{5+2\text{x}}-\big(\sqrt{3}+\sqrt{2}\big)}{\text{x}^2-6}$$=\lim\limits_{\text{x}\rightarrow{\sqrt{6}}}\frac{\sqrt{5+2\text{x}}-\Big(\sqrt{\big(\sqrt{3}+\sqrt{2}\big)^2}\Big)}{\text{x}^2-6}$
$=\lim\limits_{\text{x}\rightarrow{\sqrt{6}}}\frac{\sqrt{5+2\text{x}}-\big(\sqrt{5+2\sqrt{6}}\big)}{\text{x}^2-6}$
$=\lim\limits_{\text{x}\rightarrow{\sqrt{6}}}\frac{\sqrt{5+2\text{x}}-\big(\sqrt{5+2\sqrt{6}}\big)}{\text{x}^2-6}\times\frac{\sqrt{5+2\text{x}}+\big(\sqrt{5+2\sqrt{6}}\big)}{\sqrt{5+2\text{x}}+\big(\sqrt{5+2\sqrt{6}}\big)}$
$=\lim\limits_{\text{x}\rightarrow{\sqrt{10}}}\frac{{7+2\text{x}}-{7-2\sqrt{10}}}{\big(\text{x}^2-10\big)\Big(\sqrt{7+2\text{x}}+\big(\sqrt{7+2\sqrt{10}}\big)\Big)}$
$=\lim\limits_{\text{x}\rightarrow{\sqrt{6}}}\frac{5+2\text{x}-5-2\sqrt{6}}{\big(\text{x}^2-6\big)\Big(\sqrt{5+2\text{x}}+\big(\sqrt{5+2\sqrt{6}}\big)\Big)}$
$=\lim\limits_{\text{x}\rightarrow{\sqrt{6}}}\frac{2\big(\text{x}-\sqrt{6}\big)}{\big(\text{x}^2-\sqrt{6}\big)\Big(\sqrt{5+2\text{x}}+\big(\sqrt{5+2\sqrt{10}}\big)\Big)}$
$=\lim\limits_{\text{x}\rightarrow{\sqrt{6}}}\frac{2}{\big(\text{x}+\sqrt{6}\big)\Big(\sqrt{5+2\text{x}}+\big(\sqrt{5+2\sqrt{6}}\big)\Big)}$
$=\frac{2}{\big(2\sqrt{6}\big)\Big(2\sqrt{5+2\sqrt{6}}\Big)}$
$=\frac{1}{\big(2\sqrt{6}\big)\Big(\sqrt{5+2\sqrt{6}}\Big)}$
$=\frac{1}{\big(2\sqrt{6}\big)\Big(\sqrt{3}+\sqrt{2}\Big)}$
$=\frac{\big(\sqrt{3}-\sqrt{2}\big)}{\big(2\sqrt{6}\big)}$

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