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Limits — Maths STD 11 Science — Question

CBSE BoardEnglish MediumSTD 11 ScienceMathsLimits5 Marks
Question
Evaluate the following limit:
$\lim\limits_{\text{x}\rightarrow0}\frac{\sqrt{1+\text{x}^2}-\sqrt{1+\text{x}}}{\sqrt{1+\text{x}^3}-\sqrt{1+\text{x}}}$
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Answer

$\lim\limits_{\text{x}\rightarrow0}\frac{\sqrt{1+\text{x}^2}-\sqrt{1+\text{x}}}{\sqrt{1+\text{x}^3}-\sqrt{1+\text{x}}}$$=\lim\limits_{\text{x}\rightarrow0}\frac{\big(\sqrt{1+\text{x}^2}-\sqrt{1+\text{x}}\big)}{\big(\sqrt{1+\text{x}^2}-\sqrt{1+\text{x}}\big)}\times\frac{\big(\sqrt{1+\text{x}^2}+\sqrt{1+\text{x}}\big)}{\big(\sqrt{1+\text{x}^3}-\sqrt{1+\text{x}}\big)}$
$=\lim\limits_{\text{x}\rightarrow0}\frac{\big(1+\text{x}^2\big)-\big(1+\text{x}\big)\times\big(\sqrt{1+\text{x}^3}+\sqrt{1+\text{x}}\big)}{\big(\sqrt{1+\text{x}^3}-\sqrt{1+\text{x}}\big)\big(\sqrt{1+\text{x}^2}+\sqrt{1+\text{x}}\big)\big(\sqrt{1+\text{x}^3}+\sqrt{1+\text{x}}\big)}$
$=\lim\limits_{\text{x}\rightarrow0}\frac{\big(\text{x}^2-\text{x}\big)\big(\sqrt{1+\text{x}^3}+\sqrt{1+\text{x}}\big)}{\big(\sqrt{1+\text{x}^2}+\sqrt{1+\text{x}}\big)\big(1+\text{x}^3-1-\text{x}\big)}$
$=\lim\limits_{\text{x}\rightarrow0}\frac{\text{x}(\text{x}-1)\big(\sqrt{1+\text{x}^3}+\sqrt{1+\text{x}}\big)}{\big(\sqrt{1+\text{x}^2}+\sqrt{1+\text{x}}\big)\times\big(\text{x}^2-1\big)}$
$=\lim\limits_{\text{x}\rightarrow0}\frac{\text{x}(\text{x}-1)\big(\sqrt{1+\text{x}^3}+\sqrt{1+\text{x}}\big)}{\big(\sqrt{1+\text{x}^2}+\sqrt{1+\text{x}}\big)(\text{x})(\text{x}-1)(\text{x}+1)}$
$=\lim\limits_{\text{x}\rightarrow0}\frac{\big(\sqrt{1+\text{x}^3}+\sqrt{1+\text{x}}\big)}{\big(\big(\sqrt{1+\text{x}^2}+\sqrt{1+\text{x}}\big)(\text{x}+1)\big)}$
$=\frac{2}{2}=1$

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