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Find the limit $ \lim_{x\to 0} \frac{\ln(\tan(\pi/4 + ax))}{\sin(bx)} $ where b and a are two coefficients different from zero. I tried L'Hôpital,but nothing..

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    by nothing I mean, the answer should be 2,and I dont get a neat solution ...2012-11-13

2 Answers 2

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I would think that L'Hopital should work for you. The derivative of the numerator is $ \frac{1}{\tan(\pi/4 + ax)}\sec^2(\pi/4 + ax)a. $ The derivative of the denominator is $ \cos(bx)b $.

Ok, so more details: So by L'Hopitals rule the limit is: $ \lim_{x\to 0} \frac{\sec^2(\pi/4 + ax)a}{\tan(\pi/4 + ax)\cos(bx)b} = \frac{a\sec^2(\pi/4)}{b\tan(\pi/4)}. $ Now all that is left is for you to evaluate this expression.

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    I had already done it ,master,and the result was 2a/b but in my book its just 2..i was just joking..thanks anyway :D2012-11-13
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Since the following holds for $x\to0$

$\displaystyle \sin(bx)\sim bx,\quad \ln\big(\tan(\pi/4+ax)\big)=\ln\bigg(1+\frac{2\tan(ax)}{1-\tan(ax)}\bigg)\sim \frac{2\tan(ax)}{1-\tan(ax)} \sim \frac{2ax}{1-\tan(ax)}$

Hence $\displaystyle\lim_{x\to0}\frac{\ln\big(\tan(\pi/4+ax)\big)}{\sin(bx)}=\lim_{x\to0}\frac{2ax}{bx\big(1-\tan(ax)\big)}=\frac{2a}{b}$