- #1

- 21

- 0

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?