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प्रश्न
\[\lim_{x \to 1} \frac{x - 1}{\sqrt{x^2 + 3 - 2}}\]
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उत्तर
\[\lim_{x \to 1} \left[ \frac{x - 1}{\sqrt{x^2 + 3} - 2} \right]\] It is of the form \[\frac{0}{0}\]
Rationalising the denominator:
\[\lim_{x \to 1} \left[ \frac{\left( x - 1 \right)\left( \sqrt{x^2 + 3} + 2 \right)}{\left( \sqrt{x^2 + 3} - 2 \right)\left( \sqrt{x^2 + 3} + 2 \right)} \right]\]
= \[\lim_{x \to 1} \left[ \frac{\left( x - 1 \right)\left( \sqrt{x^2 + 3} + 2 \right)}{x^2 + 3 - 4} \right]\]
= \[\lim_{x \to 1} \left[ \frac{\left( x - 1 \right)\left( \sqrt{x^2 + 3} + 2 \right)}{\left( x^2 - 1 \right)} \right]\] = \[\lim_{x \to 1} \left[ \frac{\left( x - 1 \right)\left( \sqrt{x^2 + 3} + 2 \right)}{\left( x - 1 \right)\left( x + 1 \right)} \right]\]
= \[\frac{\sqrt{1 + 3} + 2}{1 + 1}\]
= \[\frac{4}{2}\]
= 2
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