How to calculate this limit by steps

mac321

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I need help with the limit:
lim(x->pi/2) (e^(sin2x) - e^(tg2x))/ ln(2x/pi)
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Yes, "tg" is not very common but is used to mean "tangent".

The first thing I would do is set \(\displaystyle x=\frac{\pi}{2}\). Then \(\displaystyle 2x= \pi\) so that both \(\displaystyle sin(2x)\) and \(\displaystyle tan(2x)\) are 0. Also \(\displaystyle \frac{2x}{\pi}= \frac{\pi}{\pi}=1\) so \(\displaystyle ln\left(2x/\pi\right)= ln(1)= 0\). So \(\displaystyle \frac{e^{sin(2x)}- e^{tan(2x)}}{ln(2x/\pi)}\) evaluated at \(\displaystyle \pi/2\) gives \(\displaystyle \frac{0}{0}\) an "undetermined" fraction. We can apply L'Hopital's rule. That is, differentiate the numerator and denominator separately. The derivative of \(\displaystyle e^{sin(2x)}- e^{tan(2x)}\) is \(\displaystyle 2 cos(2x)e^{sin(2x)}- 2 sec^2(2x)e^{tan(2x)}\) which at \(\displaystyle x= \frac{\pi}{2}\) is equal to -2+ 2= 0. The derivative of \(\displaystyle ln(2x/\pi)= ln(x)+ ln(2)- ln(\pi)\) is \(\displaystyle \frac{1}{x}\) which at \(\displaystyle x= \frac{\pi}{2}\) is \(\displaystyle \frac{2}{\pi}\).

By L'Hopital's rule, then, the limit is \(\displaystyle \frac{0}{\frac{2}{\pi}}= 0\).

(http://tutorial.math.lamar.edu/Classes/CalcI/LHospitalsR)
 
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