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Limits — MATHS STD 11 Science — Question

Rajasthan BoardEnglish MediumSTD 11 ScienceMATHSLimits3 Marks
Question
Evaluate the following limit:
$\lim\limits_{\text{x}\rightarrow\sqrt{2}}\frac{\text{x}^2-2}{{\text{x}^2+\sqrt{2}\text{x}-4}}$
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Answer

$\lim\limits_{\text{x}\rightarrow\sqrt{2}}\frac{\text{x}^2-2}{{\text{x}^2+\sqrt{2}\text{x}-4}}$
$=\lim\limits_{\text{x}\rightarrow\sqrt{2}}\frac{\big(\text{x}-\sqrt{2}\big)\big(\text{x}+\sqrt{2}\big)}{\text{x}^2+2\sqrt{2}\text{x}-\sqrt{2}\text{x}-4}$
$=\lim\limits_{\text{x}\rightarrow\sqrt{2}}\frac{\big(\text{x}-\sqrt{2}\big)\big(\text{x}+\sqrt{2}\big)}{\text{x}\big(\text{x}+2\sqrt{2}\big)-\sqrt{2\big(\text{x}+2\sqrt{2}\big)}}$
$=\lim\limits_{\text{x}\rightarrow\sqrt{2}}\frac{\big(\text{x}-\sqrt{2}\big)\big(\text{x}+\sqrt{2}\big)}{\big(\text{x}+2\sqrt{2}\big)\big(\text{x}-\sqrt{2}\big)}$
$=\frac{\sqrt{2}+\sqrt{2}}{\sqrt{2}+2\sqrt{2}}=\frac{2\sqrt{2}}{3\sqrt{2}}$
$=\frac23$

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