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Limits — MATHS STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceMATHSLimits4 Marks
Question
Evaluate the following limit: $\lim\limits_{\text{x}\rightarrow0}\frac{\sqrt{1+\text{x}+\text{x}^2}-\sqrt{\text{x}+1}}{2\text{x}^2}$
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Answer

$\lim\limits_{\text{x}\rightarrow0}\frac{\sqrt{1+\text{x}+\text{x}^2}-\sqrt{\text{x}+1}}{2\text{x}^2}$ $\lim\limits_{\text{x}\rightarrow0}\frac{\big(\sqrt{1+\text{x}+\text{x}^2}-\sqrt{\text{x}+1}\big)\times\big(\sqrt{1+\text{x}+\text{x}^2}+\sqrt{\text{x}+1}\big)}{2\text{x}^2\big(\sqrt{1+\text{x}+\text{x}^2}+\sqrt{\text{x}+1}\big)}$ $\lim\limits_{\text{x}\rightarrow0}\frac{\big(1+\text{x}+\text{x}^2\big)-(\text{x}+1)}{2\text{x}^2\big(\sqrt{1+\text{x}+\text{x}^2}+\sqrt{\text{x}+1}\big)}$ $\lim\limits_{\text{x}\rightarrow0}\frac{\text{x}^2}{2\text{x}^2\big(\sqrt{1+\text{x}+\text{x}^2}+\sqrt{\text{x}+1}\big)}$ $=\frac{1}{2\big(\sqrt{1}+\sqrt{1}\big)}$ $=\frac{1}{2\times2}$ $=\frac14$

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