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प्रश्न
\[\lim_{n \to \infty} \frac{1 - 2 + 3 - 4 + 5 - 6 + . . . . + \left( 2n - 1 \right) - 2n}{\sqrt{n^2 + 1} + \sqrt{n^2 - 1}}\] is equal to
पर्याय
\[\frac{1}{2}\]
\[- \frac{1}{2}\]
1
−1
उत्तर
\[- \frac{1}{2}\]
\[\lim_{n \to \infty} \left[ \frac{1 - 2 + 3 - 4 + 5 - 6 + . . . \left( 2n - 1 \right) - 2n}{\sqrt{n^2 + 1} + \sqrt{n^2 - 1}} \right]\]
\[ = \lim_{n \to \infty} \left[ \frac{\left( 1 + 3 + 5 + . . . 2n - 1 \right) - \left( 2 + 4 + 6 + . . . 2n \right)}{\left( \sqrt{n^2 + 1} + \sqrt{n^2 - 1} \right)} \right]\]
\[ = \lim_{n \to \infty} \left[ \frac{\frac{n}{2}\left( 1 + 2n - 1 \right) - \frac{n}{2}\left( 2 + 2n \right)}{\left( \sqrt{n^2 + 1} + \sqrt{n^2 - 1} \right)} \right]\]
\[ = \lim_{n \to \infty} \left[ \frac{n^2 - n\left( n + 1 \right)}{\left( \sqrt{n^2 + 1} + \sqrt{n^2 - 1} \right)} \right]\]
\[ = \lim_{n \to \infty} \left[ \frac{- n}{\left( \sqrt{n^2 + 1} + \sqrt{n^2 - 1} \right)} \right]\]
Dividing the numerator and the denominator by n:
\[= \lim_{n \to \infty} \left[ \frac{- 1}{\sqrt{1 + \frac{1}{n^2}} + \sqrt{1 - \frac{1}{n^2}}} \right] \]
\[ = \frac{- 1}{1 + 1}\]
\[ = \frac{- 1}{2}\]
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