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AP EAMCET201824 Apr 2018Morning ShiftMathematicsLimitsActual

_ n [ (1+ 1 n^2 ) (1+ 2^2 n^2 ) (1+ n^2 n^2 ) ]^ 1 n =

Options

  1. A3 e^ -4 6
  2. B2 e^ -2 4
  3. C2 e^ -4 2
  4. D4 e^ -4 4

Correct answer

C. 2 e^ -4 2

Step-by-step solution

Let A= _ n [ (1+ 1 n^2 ) (1+ 2^2 n^2 ) (1+ n^2 n^2 ) ]^ 1 n taking log on both side gathered _e A= _ n 1 n [ (1+ 1 n^2 )+ (1+ 2^2 n^2 ) (1+ n^2 n^2 ) ] A= _ n _ r=1 ^n (1+ r^2 n^2 ) 1 n = ₀^1 (1+x^2 ) d x gathered [Applying formula _ n _ r=1 ^n f ( r n ) 1 n = ₀^1 f(x) d x = ₀^5 (1+x^2 ) 1 d x using by parts, aligned & = (1+x^2 ) 1 d x- [ d d x ( (1+x^2 ) 1 d x ] d x . & =x (1+x^2 )- 2 x 1+x^2 x d x & =x (1+x^2 )-2 x^2 1+x^2 d x & =x (1+x^2 )-2 1+x^2 1+x^2 +2 1 1+x^2 d x & =x (1+x^2 )-2 1 d x+2 ⁻¹(x) & = [x (1+x^2

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