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Limits — Maths STD 11 Science — Question

Maharashtra BoardEnglish MediumSTD 11 ScienceMathsLimits4 Marks
Question
Evaluate the following limit:
$\lim\limits_{\text{x}\rightarrow{\sqrt{2}}}\frac{\sqrt{5+2\text{x}}-\big(\sqrt{2}+1\big)}{\text{x}^2-2}$
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Answer

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

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