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प्रश्न
\[\lim_{x \to - \infty} \left( \sqrt{4 x^2 - 7x} + 2x \right)\]
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उत्तर १
\[\lim_{x \to - \infty} \left( \sqrt{4 x^2 - 7x} + 2x \right)\]
Let x =\[-\] m When n → – ∞, then m → ∞.
\[ = \lim_{m \to \infty} \left[ \left( \sqrt{4 m^2 = 7m} - 2m \right) \times \frac{\left( \sqrt{4 m^2 + 7m} + 2m \right)}{\left( \sqrt{4 m^2 + 7m} + 2m \right)} \right]\]
\[ = \lim_{m \to \infty} \left[ \frac{\left( 4 m^2 + 7m \right) - \left( 2m \right)^2}{\sqrt{4 m^2 + 7m} + 2m} \right]\]
\[ = \lim_{m \to \infty} \left[ \frac{4 m^2 + 7m - 4 m^2}{\sqrt{4 m^2 + 7m} + 2m} \right]\]
Dividing the numerator and the denominator by m:
\[\lim_{m \to \infty} \left[ \frac{7}{\sqrt{\frac{4 m^2 + 7m}{m^2}} + \frac{2m}{m}} \right]\]
\[ = \lim_{m \to \infty} \left[ \frac{7}{\sqrt{\frac{4 m^2}{m^2} + \frac{7m}{m^2}} + 2} \right]\]
\[ = \lim_{m \to \infty} \left[ \frac{7}{\sqrt{4 + \frac{7}{m}} + 2} \right]\]
\[\text{ As } m \to \infty , \frac{1}{m} \to 0\]
\[ = \frac{7}{\sqrt{4} + 2}\]
\[ = \frac{7}{4}\]
उत्तर २
\[\lim_{x \to - \infty} \left( \sqrt{4 x^2 - 7x} + 2x \right)\]
Let x =\[-\] m When n → – ∞, then m → ∞.
\[ = \lim_{m \to \infty} \left[ \left( \sqrt{4 m^2 = 7m} - 2m \right) \times \frac{\left( \sqrt{4 m^2 + 7m} + 2m \right)}{\left( \sqrt{4 m^2 + 7m} + 2m \right)} \right]\]
\[ = \lim_{m \to \infty} \left[ \frac{\left( 4 m^2 + 7m \right) - \left( 2m \right)^2}{\sqrt{4 m^2 + 7m} + 2m} \right]\]
\[ = \lim_{m \to \infty} \left[ \frac{4 m^2 + 7m - 4 m^2}{\sqrt{4 m^2 + 7m} + 2m} \right]\]
Dividing the numerator and the denominator by m:
\[\lim_{m \to \infty} \left[ \frac{7}{\sqrt{\frac{4 m^2 + 7m}{m^2}} + \frac{2m}{m}} \right]\]
\[ = \lim_{m \to \infty} \left[ \frac{7}{\sqrt{\frac{4 m^2}{m^2} + \frac{7m}{m^2}} + 2} \right]\]
\[ = \lim_{m \to \infty} \left[ \frac{7}{\sqrt{4 + \frac{7}{m}} + 2} \right]\]
\[\text{ As } m \to \infty , \frac{1}{m} \to 0\]
\[ = \frac{7}{\sqrt{4} + 2}\]
\[ = \frac{7}{4}\]
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