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 October 19th, 2014, 08:37 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, 04:42 PM #2 Global Moderator   Joined: May 2007 Posts: 6,683 Thanks: 658 You can't use Taylor series around x = 0, ln(10x) has a singularity there. Thanks from topsquark
 October 20th, 2014, 04:38 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: $\displaystyle \lim_{x \to 0} \frac{\ln (ax)}{\frac{1}{\arctan (bx)}} = -\frac{1}{b} \lim_{x \to 0} \frac{\arctan^2 (bx)}{x} = -\frac{1}{b}\lim_{x \to 0} 2\arctan (bx) \frac{b}{1 + b^2x^2} = 0$. Thanks from uint

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