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प्रश्न
Write the value of \[\lim_{x \to - \infty} \left( 3x + \sqrt{9 x^2 - x} \right) .\]
उत्तर
\[\lim_{x \to - \infty} \left( 3x + \sqrt{9 x^2 - x} \right)\]
\[\text{ Let } \]
\[m = - x\]
\[\text{ If } x \to - \infty , \text{ then m } \to \infty . \]
\[ = \lim_{m \to \infty} \left( - 3m + \sqrt{9 m^2 + m} \right)\]
\[\text{ Rationalising the numerator, we get }: \]
\[ = \lim_{m \to \infty} \frac{\left( - 3m + \sqrt{9 m^2 + m} \right) \left( - 3m - \sqrt{9 m^2 + m} \right)}{- 3m - \sqrt{9 m^2 + m}}\]
\[ = \lim_{m \to \infty} \left( \frac{- \left[ - 9 m^2 + \left( 9 m^2 + m \right) \right]}{- 3m - \sqrt{9 m^2 + m}} \right)\]
\[ = \lim_{m \to \infty} \left( \frac{- m}{- 3m - \sqrt{9 m^2 + m}} \right)\]
\[\text{ Dividing the numerator and the denominator bym, and applying limit, we get }:\]
\[ = \frac{- 1}{- 3 - \sqrt{9 + \frac{1}{\infty}}}\]
\[ = \frac{1}{6}\]
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