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Limits — Maths STD 11 Science — Question

Maharashtra BoardEnglish MediumSTD 11 ScienceMathsLimits4 Marks
Question
Evaluate the following limit:
$\lim\limits_{\text{x}\rightarrow3}\frac{\text{x}-3}{\sqrt{\text{x}-2}-\sqrt{4-\text{x}}}$
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Answer

$\lim\limits_{\text{x}\rightarrow3}\frac{\text{x}-3}{\sqrt{\text{x}-2}-\sqrt{4-\text{x}}}$$=\lim\limits_{\text{x}\rightarrow3}\frac{\text{x}-3}{\big(\sqrt{\text{x}-2}-\sqrt{4-\text{x}}\big)}\times\frac{\big(\sqrt{\text{x}}-2+\sqrt{4-\text{x}}\big)}{\big(\sqrt{\text{x}}-2+\sqrt{4-\text{x}}\big)}$
$=\lim\limits_{\text{x}\rightarrow3}\frac{(\text{x}-3)\big(\sqrt{\text{x}-2}+\sqrt{4-\text{x}}\big)}{(\text{x}-2)-(4-\text{x})}$
$=\lim\limits_{\text{x}\rightarrow3}\frac{(\text{x}-3)\big(\sqrt{\text{x}-2}+\sqrt{4-\text{x}}\big)}{\text{x}-2-4+\text{x}}$
$=\lim\limits_{\text{x}\rightarrow3}\frac{(\text{x}-3)\big(\sqrt{\text{x}-2}+\sqrt{4-\text{x}}\big)}{2(\text{x}-3)}$
$=\frac{1}{2}\lim\limits_{\text{x}\rightarrow3}\big(\sqrt{\text{x}-2}+\sqrt{4-\text{x}}\big)$
$=\frac{1}{2}\big(\sqrt{3-2}+\sqrt{4-\text{x}}\big)$
$=\frac{1}{2}\big(\sqrt{1}+\sqrt{1}\big)$
$=\frac{1}{2}(1+1)=\frac22$
$=1$

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