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..
Limit question math here please?
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0by nothing I mean, the answer should be 2,and I dont get a neat solution ... – 2012-11-13
2 Answers
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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0I had already done it ,master,and the result was 2a/b but in my book its just 2..i was just joking..thanks anyway :D – 2012-11-13
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 $\displaystyle\lim_{x\to0}\frac{\ln\big(\tan(\pi/4+ax)\big)}{\sin(bx)}=\lim_{x\to0}\frac{2ax}{bx\big(1-\tan(ax)\big)}=\frac{2a}{b}$