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Limits — MATHS STD 11 Science — Question

Rajasthan BoardEnglish MediumSTD 11 ScienceMATHSLimits3 Marks
Question
Evaluate the following limit:
$\lim\limits_{\text{x}\rightarrow3}\frac{\sqrt{\text{x}+3}-\sqrt{6}}{\text{x}^2-9}$
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Answer

$\lim\limits_{\text{x}\rightarrow3}\frac{\sqrt{\text{x}+3}-\sqrt{6}}{\text{x}^2-9}$
$=\lim\limits_{\text{x}\rightarrow3}\frac{\big(\sqrt{\text{x}+3}-\sqrt{6}\big)\big(\sqrt{\text{x}+3}-\sqrt{6}\big)}{(\text{x}-3)(\text{x}+3)\big(\sqrt{\text{x}+3}+\sqrt{6}\big)}$
$=\lim\limits_{\text{x}\rightarrow3}\frac{((\text{x}+3)-6)}{(\text{x}-3)(\text{x}+3)\big(\sqrt{\text{x}+3}+\sqrt{6}\big)}$
$=\lim\limits_{\text{x}\rightarrow3}\frac{(\text{x}-3)}{(\text{x}-3)(\text{x}+3)\big(\sqrt{\text{x}+3}+\sqrt{6}\big)}$
$=\lim\limits_{\text{x}\rightarrow3}\frac{1}{(\text{x}+3)\big(\sqrt{\text{x}+3}+\sqrt{6}\big)}$
$=\frac{1}{(3+3)\sqrt{3+3}+\sqrt{6}}$
$=\frac{1}{6\big(\sqrt{6}+\sqrt{6}\big)}=\frac{1}{6\times2\sqrt{6}}$
$=\frac{1}{12\sqrt{6}}$

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