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Differential Calculus
30 Dec 2008 22:59:54 IST
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31 Dec 2008 15:18:45 IST
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I cannot understand the third question. It is not clear.
If you are asking to find the limit of (r^3-8)/(r^3+8), when r approaches infinity, the answer is 1 because
(r^3-8)/(r^3+8) = (1-8/r^3)/(1+8/r^3) = (1+0)/(1-0) = 1
If it is not this, please make the question clear first.
If you are asking to find the limit of (r^3-8)/(r^3+8), when r approaches infinity, the answer is 1 because
(r^3-8)/(r^3+8) = (1-8/r^3)/(1+8/r^3) = (1+0)/(1-0) = 1
If it is not this, please make the question clear first.
I guess
31 Dec 2008 18:39:40 IST
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3 ( r^3-8)/(r^3 + 8)
= {( r-2)/(r+2)} { (r^2 + 2r + 4)/( r^2-2r +4)}
= [ (r-2)/{r-2) +4}] * [ (r^2+2r + 4)/{(r-2)^2 + 2(r-2) +4}]
= Ar * Br (say)
Now put r= 3,4 .... ad inf And compute the product Ar and Br seperately
It is clear that , the product written in the above form , allows terms to cancel in Nr and Dr. after a period of 4 and 2 respectively for Ar and Br.
Now calculate it , answer = 1
1 Jan 2009 15:39:52 IST
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If it is r=3 to infinity, ayush, i am getting the answer as 2/7.
Solution:
Any term with r=k
=(k^3-8)/(k^3+8)=
(k-2)(k^2 + 2k + 4) / (k+2)(k^2 - 2k +4)=
(k-2)[(k + 1)^2 + 3] / (k+2)[(k - 1)^2 + 3].
Now write all the terms from k=3 to k=7
You will see that (3+2) cancels (7-2), (4+2) cancels (8-2)... and so on.
[(3+1)^2+3] cancels [(5-1)^2+3],[(4+1)^2+3] cancels [(6-1)^2+3]..... and so on.
The only terms remaining are (3-2)(4-2)(5-2)(6-2)/[(3-1)^2+3][(4-1)^2+3] = 1.2.3.4/7.12 = 2/7.
Solution:
Any term with r=k
=(k^3-8)/(k^3+8)=
(k-2)(k^2 + 2k + 4) / (k+2)(k^2 - 2k +4)=
(k-2)[(k + 1)^2 + 3] / (k+2)[(k - 1)^2 + 3].
Now write all the terms from k=3 to k=7
You will see that (3+2) cancels (7-2), (4+2) cancels (8-2)... and so on.
[(3+1)^2+3] cancels [(5-1)^2+3],[(4+1)^2+3] cancels [(6-1)^2+3]..... and so on.
The only terms remaining are (3-2)(4-2)(5-2)(6-2)/[(3-1)^2+3][(4-1)^2+3] = 1.2.3.4/7.12 = 2/7.
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1.
[(x+a)(x+b)(x+c)(x+d) ]^1/4 -x
=x [ {(1 + a/x)(1 + b/x)(1+c/x)(1+d/x)}^1/4 - 1 ]
=x [ {(1 + a/x)^x (1 + b/x)^x (1+c/x)^x (1+d/x)^x}^1/4x - 1 ]
=x [ {e^a*e^b*e^c*e^d}^1/4 - 1 ] As (1+a/x)^x={(1+a/x)^x/a}^a=e^a when x app. infinity.
=x [ {e^(a+b+c+d)/4x} - 1 ]
= x{ 1+(a+b+c+d)/4x -1} As e^y = 1+y when y app. 0.
= (a+b+c+d)/4
2. These two relations are only true when 0<x<pi/2
Proof:
sin 0=0.
Now d sin x/dx=cos x<1=dx/dx when 0<x<pi/2.
As sin x and x are both positive(and differentiable) when
0<x<pi/2
d tan x/dx=sec^2 x>1=dx/dx when 0<x<pi/2
As tan x and x are both positive(and differentiable) when
0<x<pi/2 and d tan x/dx> dx/dx, tan x>x, or tan x