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Limits — Maths STD 11 Science — Question

Maharashtra BoardEnglish MediumSTD 11 ScienceMathsLimits3 Marks
Question
Evaluate the following limit:
$\lim\limits_{\text{x}\rightarrow0}\frac{\sqrt{1+\text{x}+\text{x}^2}-1}{\text{x}}$
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Answer

$\lim\limits_{\text{x}\rightarrow0}\frac{\sqrt{1+\text{x}+\text{x}^2}-1}{\text{x}}$$=\lim\limits_{\text{x}\rightarrow0}\frac{\big(\sqrt{1+\text{x}+\text{x}^2}-1\big)}{\text{x}}\frac{\big(\sqrt{1+\text{x}+\text{x}^2}+1\big)}{\big(\sqrt{1+\text{x}+\text{x}^2}+1\big)}$
$=\lim\limits_{\text{x}\rightarrow0}\frac{\big(\big(1+\text{x}+\text{x}^2\big)-1\big)}{\big(\sqrt{1+\text{x}+\text{x}^2}+1\big)}$
$=\lim\limits_{\text{x}\rightarrow0}\frac{\text{x}(1+\text{x})}{\big(\sqrt{1+\text{x}+\text{x}^2}+1\big)}$
$=\lim\limits_{\text{x}\rightarrow0}\frac{1+\text{x}}{\sqrt{1+\text{x}+\text{x}^2}+1}$
$=\frac{1+0}{\sqrt{1+0+0}+1}$
$=\frac{1}{1+1}$
$=\frac12$

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