Limit and Integration Problem

[neo_maya]

lim (a^x-x^a)/(x^x-a^a) = 1

x -> a

 

Then what is the value of a?

 

 

:lint: (sinx/cosx)^1/2 dx [ Integration by parts? ]

 

 

:lint: 1/(sin^4 x +cos^4 x) dx

[neo_maya]

Hi,

 

Were these questions posted here before? If so, then I am sorry and if someone can show me the link, I would really appreciate it. But I have searched the calculus forum, didn't find anything.

 

I really need some help here.

 

Thanks.

[Kedas]

about your limit 'problem'.

 

denominator and numerator are the same if x=a --> 1 therefore it's true for all values of a.

(a=0 or infinite would be a special case though.)

 

I must say my math is rusty so it's probably not completely correct because 0/0 isn't really equal to 1.

[neo_maya]

Kedas said in post #3 :

about your limit 'problem'.

 

denominator and numerator are the same if x=a --> 1 therefore it's true for all values of a.

(a=0 or infinite would be a special case though.)

 

I must say my math is rusty so it's probably not completely correct because 0/0 isn't really equal to 1.

 

Hi,

 

Thanks for replying Kedas.

 

My hands r really tired. What I have written below is absolutely crap and has no meaning whatsoever. Please pardon me if I am wrong somewhere (if u actually have the patience to read this) and if u could kindly point that out, I will be really grateful. I am just trying to learn and check what I understand is right, nothing else.

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I am new at this kind of stuff and don't yet (and never will) have a clear conception of these (especially limit). But, I saw a rule called L'Hospital's Rule which says something like this

 

http://mathworld.wolfram.com/LHospitalsRule.html

 

Now notice the first line where it states something like this -

 

If lim [f'(x)/g'(x)] has a finite value .................................

 

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I will come to this rule later

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What I am trying to say is this - in most of the cases of limit probs - we have to compute the value of a discontinuous function (at least what I understand) which doesn't have a real value if we put the value of the limit in the place of the function's variable. Most of the time, if we simply put the limit, then we will get a bizzare form of some sort like - 0/0 or infinity/infinity etc.

 

 

Actually, the form is an indeterminate form. Meaning that you have'n determined the answer yet. This indeterminate form can be easily (?) circumvented by using algebraic manipulation. Such tools as algebraic simplification, factoring, and conjugates can easily be used to circumvent the form so that the limit can be calculated.

 

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A simple example :

 

lim {(x^3 - 27)/(x^2-9)}

x->3

 

Now - if we just put 3 replacing x then we will get a form like 0/0. Thus, we can't compute it. But actually we can. We can cancel the (x-3) term and then compute it and the result is 9/2.

 

There r lots of these examples - like when u work out the derivative of lnx. First u simplify lnx in its series (I think u get the series from maclaurine's theorem ? ) and then do some algebric manipulation and bang - u have the result (1/x).

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Now my point is that all these examples had a form like 0/0 - yet they produce a result. I think (?) what we are doing here is that we first cancel that part of the function for which we get an indeterminant form and then we compute it.

 

So, doesn't this mean that my limit problem has a solution too ? Now, look at this example

 

lim a*{(x^3 - 27)/(x^2-9)} = 1

x->3

 

shouldn't we get the value of a = 2/9?

 

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Back to the L'Hospital's Rule - if I understood the rule right and what I did was right then u get the following form of my problem :

 

( ln a - 1 ) / ( lna + 1 ) = 1

=> lna - 1 = ln +1

=> nothing

 

Here's my problem ????????

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[Kedas]

Oh yes, L'Hospital's Rule, this does ring a lot of bells (and trigger memories)

 

Yes, what I said was useless, Sorry.

This '( ln a - 1 ) / ( lna + 1 )' is what you got when you made the derivative to x of both?

 

BTW any idea about the derivative of x^x ?

[neo_maya]

Kedas said in post #5 :

Oh yes, L'Hospital's Rule, this does ring a lot of bells (and trigger memories)

 

Yes, what I said was useless, Sorry.

This '( ln a - 1 ) / ( lna + 1 )' is what you got when you made the derivative to x of both?

 

BTW any idea about the derivative of x^x ?

 

 

Lol.

 

Let,

 

x^x = y

 

=> ln(x^x) = lny

 

=>xlnx = lny

 

=>d(xlnx)/dx = dy/dx

 

=>x(d(lnx)/dx)+lnx(dx/dx) = (1/y)(dy/dx)

 

=>1+lnx = (1/x^x)(dy/dx)

 

=>x^x(1+lnx) = d(x^x)/dx

 

 

So, d(x^x)/dx = x^x(1+lnx)

 

 

EDIT :

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PS : What is the meaning of BTW?

 

I only know two of these shortcuts - LOL, ROFL.

 

 

[neo_maya]

Don't tell me, don't tell me - Is it BY THE WAY?

[Kedas]

So what you ask is:

x^x(1+lnx)=a^x . lnx - a.x^(a-1)

 

what is 'a' ? right ? hmm I have no idea now. (with a=x)

 

yes, By The Way

[neo_maya]

Hmm, niether do I.

 

But Thanks a lot for trying.

[neo_maya]

:help:

[neo_maya]

 

Hey guys, I think I have solved the last one of my three posted problems. And no need to try to solve the limit problem. A friend of mine told me that it was not 1, it was -1.

 

Don't know if any of u care but I would post the solution anyway since I think I have bugged u all enough with this. Here it is -

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:int: (1/(sin^4 x + cos^4 x) dx

 

= 4* :int: 1/ [ (2sin^2 x)^2 + (2cos^2 x)^2 ] dx

 

= 4* :int: 1/ [ (1-co2x)^2 + (1+cos2x)^2 ] dx

 

= 4* :int: 1/ [ 2+2cos^2 2x ] dx

 

= 2* :int: sec^2 2x/ [ 2 + tan^2 2x ] dx

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let,

 

tan2x = z

 

=> 2sec^2 2x dx = dz

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= :int: 1/ [ {2^(1/2)}^2 } + z^2 ] dx

 

= [1/{2^(1/2)}^2 } ] * tan(inverse) [ tan2x / {2^(1/2)}^2 } ] + c

 

 

TADA.

 

Though couldn't solve the second one. Any idea about that one ?