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 June 22nd, 2012, 01:18 PM #1 Senior Member   Joined: Jan 2012 Posts: 159 Thanks: 0 Applying L'Hopital's Rule It starts in the form 0/0 so it is of the right form to apply L'Hopital's Rule but the trig numerator I think needs a manipulation and so does the denominator where the ln is ln(x+1)^2 which goes to 2ln(x+1) but I am a bit stuck any ideas?
 June 22nd, 2012, 04:48 PM #2 Senior Member   Joined: May 2011 Posts: 501 Thanks: 5 Re: Applying L'Hopital's Rule You can use L'Hopital. But, it takes about 3 applications. You can write it as $\frac{sin(x)}{2x^{2}\ln(x+1)}-\frac{\tan^{-1}(x)}{2x^{2}\ln(x+1)}$ For the left side, applying L'H 3 times brings us to: $\frac{-cos(x)(x+1)^{3}}{4(x^{2}+3x+3)}$ Now, letting x=0, gives $\frac{-1}{12}$ For the right side, after applying L'H 3 times, we get: $\frac{2(3x^{2}-1)}{(x^{2}+1)^{3}}\cdot \frac{(x+1)^{3}}{4(x^{2}+3x+3)}$ Letting x=0, gives $\frac{-1}{6}$ Putting them together, gives $\frac{-1}{12}-(\frac{-1}{6})=\frac{1}{12}$ This rather daunting to do all of that differentiation. There is more than likely a better and more clever way. Someone will come along with a sharp method without L'H.
 June 22nd, 2012, 04:53 PM #3 Senior Member   Joined: Jan 2012 Posts: 159 Thanks: 0 Re: Applying L'Hopital's Rule thanks man going through your steps to see where I messed up!
 June 22nd, 2012, 09:01 PM #4 Senior Member   Joined: Feb 2009 From: Adelaide, Australia Posts: 1,519 Thanks: 3 Re: Applying L'Hopital's Rule Because x is approaching 0, you can immediately use the well-known series: $\sin x= x-\frac{x^3}{6}+\frac{x^5}{120}-...\quad\text{for all }x\\\arctan x =x-\frac{x^3}{3}+\frac{x^5}{5}-...\quad\text{for }x^2\le 1$ For very small x, you can discard higher powers, so the numerator is $x-\frac{x^3}{6}+...-(x-\frac{x^3}{3}+...)\approx\frac{x^3}{6}$ Meanwhile $\ln(1+u)= u-\frac12u^2+\frac13u^3-\frac14u^4+...\quad\text{for }-1

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