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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: 6,822 Thanks: 723 You can't use Taylor series around x = 0, ln(10x) has a singularity there. Thanks from topsquark 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: $\displaystyle \lim_{x \to 0} \frac{\ln (ax)}{\frac{1}{\arctan (bx)}} = -\frac{1}{b} \lim_{x \to 0} \frac{\arctan^2 (bx)}{x} = -\frac{1}{b}\lim_{x \to 0} 2\arctan (bx) \frac{b}{1 + b^2x^2} = 0$. Thanks from uint Tags series, solve, taylor Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post g0bearmon Real Analysis 2 May 22nd, 2012 12:10 PM FreaKariDunk Real Analysis 3 April 27th, 2012 08:57 PM Singularity Real Analysis 1 August 25th, 2010 07:27 PM stewpert Calculus 3 February 10th, 2010 01:15 PM g0bearmon Calculus 1 December 31st, 1969 04:00 PM

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