- #1
Bazman
- 21
- 0
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?
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?