Calculus

Question

Comments

infinity-infinity=0

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Multiply with the conjugate of $\sqrt{n^2 + 2n} - [ \sqrt{n^2 + 2n} ]$ in both numerator and denominator and put n= infinite then it will result in 0.

Hence A option should be correct.

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ya  ∞ − ∞ is indeterminate, so we rationalize the equation,and then put n, to get a 0.Sry for previous comment.

Answer 1

Multiply with the conjugate of $\sqrt{n^2 + 2n} - [ \sqrt{n^2 + 2n} ]$ in both numerator and denominator and put n= infinite then it will result in 0.

Hence A option should be correct.

Comments

can u show calculation?

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Shown