Evaluate the following limit: _ x 0 1+x+x^2 - x+1 2x^2 - Vidyadip

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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