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प्रश्न
\[\lim_{x \to 0} \frac{\sqrt{1 + x + x^2} - 1}{x}\]
उत्तर
\[\lim_{x \to 0} \left[ \frac{\sqrt{1 + x + x^2} - 1}{x} \right]\]
When x = 0, the expression \[\frac{\sqrt{1 + x + x^2} - 1}{x}\] takes the form\[\frac{0}{0}\]
Rationalising the numerator:
\[\lim_{x \to 0} \left[ \frac{\left( \sqrt{1 + x + x^2} - 1 \right)\left( \sqrt{1 + x + x^2} + 1 \right)}{x\left( \sqrt{1 + x + x^2} + 1 \right)} \right]\]
\[ = \lim_{x \to 0} \left[ \frac{1 + x + x^2 - 1}{x\left( \sqrt{1 + x + x^2} + 1 \right)} \right]\]
\[ = \lim_{x \to 0} \left[ \frac{x\left( 1 + x \right)}{x\left( \sqrt{1 + x + x^2} + 1 \right)} \right]\]
\[ = \frac{1 + 0}{\sqrt{1 + 0 + 0} + 1}\]
\[ = \frac{1}{2}\]
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