December 23rd, 2016, 03:30 AM  #1 
Member Joined: Mar 2016 From: Sweden Posts: 31 Thanks: 4  Taylor series
I have the following problem to solve $$\lim_{x\to0} (\frac{cosx}{cos2x})^{1/x^2}$$ so I take ln and use Taylor series for cosx. For cosx I get $\mathcal{O}(x^4)$. Then I am confused when I should use the Taylor series for $ln(12x^2+\mathcal{O}(x^4))$. I am confused about the $\mathcal{O}(x^4)$ inside ln when using Taylor. Hopefully you understand my question, thanks 
December 23rd, 2016, 04:00 AM  #2 
Senior Member Joined: Sep 2007 From: USA Posts: 349 Thanks: 67 Math Focus: Calculus 
Honestly, I wouldn't use Taylor series for that limit. Is it a requirement?

December 23rd, 2016, 06:20 AM  #3 
Member Joined: Mar 2016 From: Sweden Posts: 31 Thanks: 4  
December 23rd, 2016, 06:44 AM  #4  
Math Team Joined: Dec 2013 From: Colombia Posts: 6,388 Thanks: 2100 Math Focus: Mainly analysis and algebra  Quote:
\ln(1+x) &= x\tfrac12x^2+\mathcal{O}(x^3) \\ \ln \big(12x^2+\mathcal{O}(x^4) \big) &= \big( 12x^2+\mathcal{O}(x^4) \big)  \frac12 \big(12x^2+\mathcal{O}(x^4) \big)^2 + \mathcal{O}(x^3) \end{align*}$$ and then expand noting that $a\mathcal{O}(x^n)^2=\mathcal{O}(x^{2n})$ and $ax^m\mathcal{O}(x^n)=\mathcal{O}(x^{m+n})$. Edit: and for $m \lt n$ we have $\mathcal{O}(x^m) + \mathcal{O}(x^n) = \mathcal{O}(x^{m})$. Last edited by v8archie; December 23rd, 2016 at 06:48 AM.  
January 16th, 2017, 10:03 PM  #5  
Newbie Joined: Jan 2017 From: VN Posts: 2 Thanks: 1  Quote:
$\ln \big(12x^2+\mathcal{O}(x^4) \big)= \big(2x^2+\mathcal{O}(x^4) \big) \frac12 \big(2x^2+\mathcal{O}(x^4) \big)^2+ \mathcal{O}(x^3). $ Last edited by skipjack; January 17th, 2017 at 10:33 AM.  

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