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प्रश्न
\[\lim_{x \to \infty} \frac{x}{\sqrt{4 x^2 + 1} - 1}\]
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उत्तर
\[\lim_{x \to \infty} \left[ \frac{x}{\sqrt{4 x^2 + 1} - 1} \right]\]
\[\text{ Rationalising the denominator }: \]
\[ \lim_{x \to \infty} \left[ \frac{x}{\left( \sqrt{4 x^2 + 1} - 1 \right)} \frac{\left( \sqrt{4 x^2 + 1} + 1 \right)}{\left( \sqrt{4 x^2 + 1} + 1 \right)} \right]\]
\[ = \lim_{x \to \infty} \left[ \frac{x\left( \sqrt{4 x^2 + 1} + 1 \right)}{4 x^2 + 1 - 1} \right]\]
\[ = \lim_{x \to \infty} \left[ \frac{\sqrt{4 x^2 + 1} + 1}{4x} \right]\]
\[\text{ Dividing the numerator and the denominator by } x: \]
\[ \lim_{x \to \infty} \left[ \frac{\frac{\sqrt{4 x^2 + 1}}{x} + \frac{1}{x}}{4} \right]\]
\[ = \lim_{x \to \infty} \left[ \frac{\sqrt{\frac{4 x^2 + 1}{x^2}} + \frac{1}{x}}{4} \right]\]
\[ = \lim_{x \to \infty} \left[ \frac{\sqrt{4 + \frac{1}{x^2}} + \frac{1}{x}}{4} \right]\]
\[ x \to \infty \]
\[ \therefore \frac{1}{x}, \frac{1}{x^2} \to 0\]
\[ = \frac{\sqrt{4}}{4}\]
\[ = \frac{2}{4}\]
\[ = \frac{1}{2}\]
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