11---pi rt?
Q1)lim (x+2)tan-1(x+2) - x tan-1x
(x→ ∞)
Q2) lim ((1.5)n+ [(1+0.0001)10000]n)1/n
(x→ ∞)
Here[ ] is Greatest Integer Function (GIF)
n
Q3) lim Σ log( 1 +K/n)1/n
(n→∞) k=1
Thnx in advance
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8 Answers
12) using (1+x)n=1+nx,
expression=((3/2)n+(2)n)1/n=1/2((3/4)n+1)1/n= 1/2
3)e∫ln(1+x) dx=4 ,with 0 and 1 as limits of integration.
62(x+2)tan^{-1}(x+2) - x tan^{-1}x \\x(tan^{-1}(x+2)-tan^{-1}x)+2tan^{-1}(x+2) \\xtan^{-1}(2/(1+2x+x^2))+\pi \\\text{apply limit you will get } \pi
11 > Let , f ( y ) = y tan - 1 ( y ) .
Apply Langrange ' s Mean Value Theorem on " f " in the interval " ( x , x + 2 ) " .
For an arbitrary " c " in the said interval , we must have : -
( x + 2 ) tan - 1 ( x + 2 ) - x tan - 1 ( x ) x + 2 - x = tan - 1 ( c ) + c1 + c 2
Or , ( x + 2 ) tan - 1 ( x + 2 ) - x tan - 1 ( x ) = 2 tan - 1 ( c ) + 2 c1 + c 2 .................... ( 1 )
Now , as " x " approaches infinity , we see that both " x " and " x + 2 " tend towards infinity .
Accordingly , " c " must also tend towards infinity as it is crunched in between them .
The given limit is the R . H . S of equation ( 1 ) as " x → ∞ " , whose value is determined by letting " c " approach infinity as well and calculating the limit of L . H . S in that case .
We see that , if " c → ∞ , c1 + c 2 → 0 " , and " tan - 1 ( c ) → π2 " .
Hence , the given limit = 2 . π2 = π
71Bas ek doubt hai..
How would I know whether I can use this method to solve this problem. Actually, when we first look at the sum how do we think of reaching up the solution in this way . (Not particularly for this one, but for any typical sum)
262@samagra- there is a slight mistake.... answer of Q2. is '2' and not '1/2'