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Limits — MATHS STD 11 Science — Question

Rajasthan BoardEnglish MediumSTD 11 ScienceMATHSLimits3 Marks
Question
Evaluate the following limit:
$\lim\limits_{\text{x}\rightarrow\sqrt{3}}\frac{\text{x}^4-9}{{\text{x}^2+4\sqrt{3}\text{x}-15}}$
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Answer

$\lim\limits_{\text{x}\rightarrow\sqrt{3}}\frac{\text{x}^4-9}{{\text{x}^2+4\sqrt{3}\text{x}-15}}$
$=\lim\limits_{\text{x}\rightarrow\sqrt{3}}\frac{\big(\text{x}-\sqrt{3}\big)\big(\text{x}+\sqrt{3}\big)\big(\text{x}^2+3\big)}{{\big(\text{x}-\sqrt{3}\big)\big(\text{x}+5\sqrt{3}\big)}}$
$=\lim\limits_{\text{x}\rightarrow\sqrt{3}}\frac{\big(\text{x}+\sqrt{3}\big)\big(\text{x}^2+3\big)}{{\big(\text{x}+5\sqrt{3}\big)}}$
$=\frac{\big(\sqrt{3}+\sqrt{3}\big)(3+3)}{\big(\sqrt{3}+5\sqrt{3}\big)}=\frac{\big(3\sqrt{3}\big)(6)}{6\sqrt{3}}=2$

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