lim n → ∞ n ∑ r = 1 c o t − 1 ( r 2 + 3 4 ) is Options: A . &nbsp 0 B . &nbsp t a n − 1 1 C . &nbsp t a n − 1 2 D . &nbsp tan − 1 1 2

Answer: Option C : C lim n → ∞ n ∑ r = 1 t a n − 1 ( 4 4 r 2 + 3 ) = lim n → ∞ n ∑ r = 1 t a n − 1 ( 1 r 2 − 1 4 + 1 ) = lim n → ∞ n ∑ r = 1 t a n − 1 ( ( r + 1 2 ) − ( r − 1 2 ) 1 + ( r + 1 2 ) ( r − 1 2 ) ) = lim n → ∞ n ∑ r = 1 t a n − 1 { t a n − 1 ( r + 1 2 ) − t a n − 1 ( r − 1 2 ) } = lim n → ∞ { t a n − 1 ( n + 1 2 ) − t a n − 1 ( 1 2 ) } = π 2 − t a n − 1 1 2 = t a n − 1 2

limn→∞n∑r=1cot−1(r2+34) is

Discussion Forum : Limits Continuity And Differentiability

Question -

$lim n \to \infty n \sum r = 1 c o t^{- 1} (r^{2} + \frac{3}{4})$ is

Options:

A . 0

B .

t a n^{- 1} 1

C .

t a n^{- 1} 2

D .

{tan}^{- 1} \frac{1}{2}

Answer: Option C
:
C

$lim n \to \infty n \sum r = 1 t a n^{- 1} (\frac{4}{4 r^{2} + 3})$

= $lim n \to \infty n \sum r = 1 t a n^{- 1} (\frac{1}{r^{2} - \frac{1}{4} + 1})$

= $lim n \to \infty n \sum r = 1 t a n^{- 1} (\frac{(r + \frac{1}{2}) - (r - \frac{1}{2})}{1 + (r + \frac{1}{2}) (r - \frac{1}{2})})$

= $lim n \to \infty n \sum r = 1 t a n^{- 1} {t a n^{- 1} (r + \frac{1}{2}) - t a n^{- 1} (r - \frac{1}{2})}$

= $lim n \to \infty {t a n^{- 1} (n + \frac{1}{2}) - t a n^{- 1} (\frac{1}{2})}$
= $\frac{π}{2} - t a n^{- 1} \frac{1}{2}$
= $t a n^{- 1} 2$

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Discussion Forum : Limits Continuity And Differentiability

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