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October 19th, 2014, 07:37 AM  #1 
Member Joined: Oct 2014 From: Norway Posts: 42 Thanks: 1  How can I solve this using Taylor series?
lim ln(10x)arctan(7x) as x > 0 I'm trying to find this using Taylor series. However, I'm stuck. If I try to make maclaurin formula out of each product separately, then the formula for ln(7x) starts with the term ln(7x)... And if I try to make a taylor series for the whole expression, I again get terms with ln(7x). Anyone know how I can successfully use taylor series to solve this? Thanks. 
October 19th, 2014, 03:42 PM  #2 
Global Moderator Joined: May 2007 Posts: 6,730 Thanks: 689 
You can't use Taylor series around x = 0, ln(10x) has a singularity there.

October 20th, 2014, 03:38 AM  #3 
Senior Member Joined: Dec 2013 From: some subspace Posts: 212 Thanks: 72 Math Focus: real analysis, vector analysis, numerical analysis, discrete mathematics 
I'd also avoid using the serie in this case... However, if I must use them, I'd write the Taylor serie of $\displaystyle \ln x$ at point $\displaystyle x = 1$ like this: $\displaystyle \lim_{x \to 0+} \ln (ax) \arctan (bx) = \lim_{x \to 0} \left[ (ax  1)  \frac{(ax  1)^2}{2} + \cdots \right] \left[ bx  \frac{(bx)^3}{3} + \cdots \right] = \lim_{x \to 0} \left( bx(ax  1) + \cdots \right) = 0$. But because there are the singularity at $\displaystyle x = 0$, I'd use e.g. l'Hopital's rule: $\displaystyle \lim_{x \to 0} \frac{\ln (ax)}{\frac{1}{\arctan (bx)}} = \frac{1}{b} \lim_{x \to 0} \frac{\arctan^2 (bx)}{x} = \frac{1}{b}\lim_{x \to 0} 2\arctan (bx) \frac{b}{1 + b^2x^2} = 0$. 

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