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Limits and Derivatives — Maths STD 11 Science — Question

CBSE BoardEnglish MediumSTD 11 ScienceMathsLimits and Derivatives3 Marks
Question
Evaluate:
$\lim\limits_{\text{x} \rightarrow \sqrt{2}}\frac{\text{x}^{2}-4}{\text{x}^{2}+3\sqrt{2\text{x}}-8}$
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Answer

Given that $\lim\limits_{\text{x} \rightarrow \sqrt{2}}\frac{\text{x}^{2}-4}{\text{x}^{2}+3\sqrt{2\text{x}}-8}$
$=\lim\limits_{\text{x} \rightarrow \sqrt{2}}\frac{(\text{x}^{2}-2)(\text{x}^{2}+2)}{\text{x}^{2}+4\sqrt{2}\text{x}-\sqrt{2}\text{x}-8}$
$=\lim\limits_{\text{x} \rightarrow \sqrt{2}}\frac{(\text{x}+\sqrt{2})(\text{x}-\sqrt{2})(\text{x}^{2}+2)}{\text{x}(\text{x}+4\sqrt{2})-\sqrt{2}(\text{x}+4\sqrt{2})}$
$=\lim\limits_{\text{x} \rightarrow \sqrt{2}}\frac{(\text{x}+\sqrt{2})(\text{x}-\sqrt{2})(\text{x}^{2}+2)}{(\text{x}+4\sqrt{2})(\text{x}-\sqrt{2})}$
$=\lim\limits_{\text{x} \rightarrow \sqrt{2}}\frac{(\text{x}+\sqrt{2})(\text{x}^{2}+2)}{\text{x}+4\sqrt{2}}$
Taking limits we have
$=\frac{(\sqrt{2}+\sqrt{2})(2+2)}{\sqrt{2}+4\sqrt{2}}$
$=\frac{2\sqrt{2}\times4}{5\sqrt{2}}=\frac{8}{5}$
Hence, required answre is $\frac{8}{5}.$

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