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Limits — MATHS STD 11 Science — Question

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

$\lim\limits_{\text{x}\rightarrow0}\frac{\text{x}}{\sqrt{1+\text{x}}-\sqrt{1-\text{x}}}$ $=\lim\limits_{\text{x}\rightarrow0}\frac{\text{x}}{\big(\sqrt{1+\text{x}}-\sqrt{1-\text{x}}\big)}\times\frac{\big(\sqrt{1+\text{x}}+\sqrt{1-\text{x}}\big)}{\big(\sqrt{1+\text{x}}+\sqrt{1-\text{x}}\big)}$ $=\lim\limits_{\text{x}\rightarrow0}\frac{\text{x}\big(\sqrt{1+\text{x}}+\sqrt{1-\text{x}}\big)}{\big(\sqrt{1+\text{x}}\big)^2-\big(\sqrt{1-\text{x}}\big)^2}$ $=\lim\limits_{\text{x}\rightarrow0}\frac{\text{x}\big(\sqrt{1+\text{x}}+\sqrt{1-\text{x}}\big)}{1+\text{x}-1+\text{x}}$ $=\lim\limits_{\text{x}\rightarrow0}\frac{\text{x}\big(\sqrt{1+\text{x}}+\sqrt{1-\text{x}}\big)}{2\text{x}}$ $=\frac12\lim\limits_{\text{x}\rightarrow0}\Big(\frac{\sqrt{1+\text{x}}+\sqrt{1-\text{x}}}{\text{x}}\Big)\text{x}$ $=\frac{1}{2}\lim\limits_{\text{x}\rightarrow0}{\big(\sqrt{1+\text{x}}+\sqrt{1-\text{x}}\big)}$ $=\frac{1}{2}\big(\sqrt{1}+\sqrt{1}\big)$ $=\frac{1}{2}(1+1)=\frac22$ $=1$

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