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