I'm not getting the algebric trick to do this limit, when

In a similar question i did the taylor expansion with 2 terms: for , and it worked fine (i that case ). Probably i'll have to use the same in here.

Thank's.

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- April 10th 2011, 12:55 PMhurzlimit when a>>x
I'm not getting the algebric trick to do this limit, when

In a similar question i did the taylor expansion with 2 terms: for , and it worked fine (i that case ). Probably i'll have to use the same in here.

Thank's. - April 10th 2011, 02:00 PMHallsofIvy
"x>> a" is the same as saying that x goes to 0 (as compared to a).

Start by rationalizing the numerator of the fraction inside the ln by multiplying both numerator and denominator by .

Now, both numerator and denominator of the argument go to 0 as x goes to 0 so we can use L'hopital's rule:differentiate both, with respect to x. The derivative of the numerator is 2x. The derivative of the denominator is . The numerator and denominator still go to 0, as x goes to 0, so we use L'Hopital again.

The derivative of 2x is 2. The derivative of [tex]2x+ 2ax(a^2+ x^2)^{-1/2}[tex] is . The first, of course, goes to 2 while the second goes to as x goes to 0. - April 10th 2011, 02:26 PMtopsquark
Without using Calculus:

Consider the numerator inside the logarithm:

(by the binomial theorem. Since a >> x we can consider only the first two terms.)

Thus

and after a little work:

So you need to calculate

for a>>x

Can you take it from here?

-Dan