Solving Basic Log Question: x^y*|ln(1/x)|^m

[Bazman]

Hi,

I come from an engineering background and so have not studied analysis (sadly). I need to figure out the following.

How does:

1.) x^y*|ln(1/x)|^m behave for any m given y<0 as x-> infinity

2.) x^y*|ln(1/x)|^m behave for any m given y>0 as x-> 0

The way I see it in the first example as x-> infinity the |ln(1/x)|-> infinity
so effectively you have infinity^y*infinity^m and y is less than 1. So this should explode right?

However the answer is apparently that the expression->x?

In the second |ln(1/x)|-> infinity as x tends to 0. So effectivey you have
infinity^m*0=0.
However the answer is apparently that the expression ->1?

Can someone please explain where I am going wrong?

 

[D H]

You can simplify this using |\log(1/x)| = |-\log x| = \log x.
Thus, x^y|\log(1/x)^m| = x^y(\log x)^m.

The following only works if m is positive. In both problems, x^y\to 0 and (\log x)^m\to\infty as x\to\infty (1) or x\to 0 (2). The product is indeterminate. Solving it calls for L'Hopital's rule.

 

[Bazman]

Hi,

I come from an engineering background and so have not studied analysis (sadly). I need to figure out the following.

How does:

1.) x^y*|ln(1/x)|^m behave for any m given y<0 as x-> infinity

2.) x^y*|ln(1/x)|^m behave for any m given y>0 as x-> 0

The way I see it in the first example as x-> infinity the |ln(1/x)|-> infinity
so effectively you have infinity^y*infinity^m and y is less than 1. So this should explode right?

However the answer is apparently that the expression->x?

In the second |ln(1/x)|-> infinity as x tends to 0. So effectivey you have
infinity^m*0=0.
However the answer is apparently that the expression ->1?

Can someone please explain where I am going wrong?

You can simplify this using |\log(1/x)| = |-\log x| = \log x.
Thus, x^y|\log(1/x)^m| = x^y(\log x)^m.

The following only works if m is positive. In both problems, x^y\to 0 and (\log x)^m\to\infty as x\to\infty (1) or x\to 0 (2). The product is indeterminate. Solving it calls for L'Hopital's rule.

Sorry problem 1 above should read

1.) x^y*|ln(1/x)|^m behave for any m given y<1 as x-> infinity

but I don;t think that changes the nature of your argument.

In any case thanks will look into L'hopital's rule