2 Answers

Let L = lim(x>0+) (tan(2x))^x.
Take ln's of both sides:
ln L = lim(x>0+) x * ln(tan(2x))
= lim(x>0+) ln(tan(2x)) / x^(1).
Since this is of the form "infinity/infinity", we can apply L'Hopital's Rule.
So,
ln L = lim(x>0+) [(1/tan(2x)) * sec^2(2x) * 2] / [x^(2)]
= lim(x>0+) (2x^2/sin(2x)) * sec(2x)
= lim(x>0+) (x) * (2x/sin(2x)) * [1/cos(2x)]
= 0 * 1/1 * 1
= 0.
Thus, L = e^0 = 1.
I hope this helps!

e
lim(x→0+)〖(tan2x)^x 〗
Y=(tan2x)^x
ln y=xln tan 2x
lim(x→0+)ln y = lim(x→0+)xlntan2x
lim(x→0+)〖lntan2x/(1/x)〗
Using L’Hopital
=lim(x→0+) ((2(sec2x)^2)/tan2x)/(1/x^2)
=2 lim(x→0+)〖x^2/(sen2x cos2x)〗
Applying for the second time
=2 lim(x→0+)〖x/((cos2x)^2(sen2x)^2)〗
0
So we find
lim ln y=0
so
lim y = e^0
lim y=1