



 
Dave, There is no surprise that you might run into trouble with your approach: It is not possible to expand your function about x = 0  as x approaches 0 from the positive side, f(x) goes to (infinity). You must therefore choose another value of x to expand about. The usual choice is x = 1, in which case the Taylor polynomial for ln x in terms of x1 is very simple to write out. (It will be a series in powers of x1.) Alternatively, you can look it up in tables. It will represent ln x for all x between 0 and 2. To get the expansion for f(x) = (ln x)/x, I would write out the expansion for 1/x about x1 (it's just the derivative of ln x), then multiply the two series together. Chris  


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