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 \(\displaystyle \frac{0}{\frac{2}{\pi}}= 0\).
This site uses cookies to help personalise content, tailor your experience and to keep you logged in if you register.
By continuing to use this site, you are consenting to our use of cookies.