मराठी

Lim X → √ 3 X 2 − 3 X 2 + 3 √ 3 X − 12

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प्रश्न

\[\lim_{x \to \sqrt{3}} \frac{x^2 - 3}{x^2 + 3 \sqrt{3}x - 12}\]

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उत्तर

\[\lim_{x \to \sqrt{3}} \left[ \frac{x^2 - 3}{x^2 + 3\sqrt{3}x - 12} \right]\]
\[\text{ It is of the form } \frac{0}{0} . \]
\[ \lim_{x \to \sqrt{3}} \left[ \frac{x^2 - \left( \sqrt{3} \right)^2}{x^2 + 4\sqrt{3}x - \sqrt{3}x - 12} \right]\]
\[ = \lim_{x \to \sqrt{3}} \left[ \frac{\left( x - \sqrt{3} \right)\left( x + \sqrt{3} \right)}{x\left( x + 4\sqrt{3} \right) - \sqrt{3}\left( x + 4\sqrt{3} \right)} \right]\]
\[ = \lim_{x \to \sqrt{3}} \left[ \frac{\left( x - \sqrt{3} \right)\left( x + \sqrt{3} \right)}{\left( x - \sqrt{3} \right)\left( x + 4\sqrt{3} \right)} \right]\]
\[ = \frac{\sqrt{3} + \sqrt{3}}{\sqrt{3} + 4\sqrt{3}}\]
\[ = \frac{2}{5}\]

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पाठ 29: Limits - Exercise 29.3 [पृष्ठ २३]

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आर.डी. शर्मा Mathematics [English] Class 11
पाठ 29 Limits
Exercise 29.3 | Q 12 | पृष्ठ २३

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