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Question
\[\lim_{x \to 0} \frac{\sqrt{1 + x + x^2} - \sqrt{x + 1}}{2 x^2}\]
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Solution
\[\lim_{x \to 0} \left[ \frac{\sqrt{1 + x + x^2} - \sqrt{x + 1}}{2 x^2} \right]\] It is of the form \[\frac{0}{0}\]
Rationalising the numerator:
\[\lim_{x \to 0} \left[ \frac{\left( \sqrt{1 + x + x^2} - \sqrt{x + 1} \right)\left( \sqrt{1 + x + x^2} + \sqrt{x + 1} \right)}{\left( \sqrt{1 + x + x^2} + \sqrt{x + 1} \right)2 x^2} \right]\]
= \[\lim_{x \to 0} \left[ \frac{\left( 1 + x + x^2 \right) - \left( x + 1 \right)}{\left( \sqrt{1 + x + x^2} + \sqrt{x + 1} \right)2 x^2} \right]\]
= \[\lim_{x \to 0} \left[ \frac{x^2}{\left( \sqrt{1 + x + x^2} + \sqrt{x + 1} \right)\left( 2 x^2 \right)} \right]\]
= \[\frac{1}{\left( \sqrt{1 + 0 + 0} + \sqrt{0 + 1} \right) \times 2}\]
= \[\frac{1}{2} \times \frac{1}{2}\]
= \[\frac{1}{4}\]
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