How to calculate this limit by steps

[mac321]

I need help with the limit:
lim(x->pi/2) (e^(sin2x) - e^(tg2x))/ ln(2x/pi)

 

[Deleted member 4993]

I need help with the limit:
lim(x->pi/2) (e^(sin2x) - e^(tg2x))/ ln(2x/pi)
View attachment 9762

Is tg2x = tan(2x)?

 

[HallsofIvy]

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)