Basic Log Question

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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?
 

Answers and Replies

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

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


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

The following only works if m is positive. In both problems, [itex]x^y\to 0[/itex] and [itex](\log x)^m\to\infty[/itex] as [itex]x\to\infty[/itex] (1) or [itex]x\to 0[/itex] (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
 

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