Question

Lt xtends to pi/2(tanx)^tan2x

Answer 1

Answer:

Let y=log(tanx) limx→π/2

Then limx→π/2log((tan(x))tan2x)=limx→π/2tan2x(logtan(x))

Using tan2x=2tan(x)/(1+tan2x))

limx→π/2tan2x(logtan(x))=limy→∞2ey1−e2yy

=2ye−y−ey=2e−yy−eyy

But

limy→∞e−yy=0

limy→∞eyy=∞ as ey/y=1/y+1+y/2!... by Taylor expansion

So

limy→∞2e−yy−eyy=0 ⟹limx→π/2log((tan(x))tan2x)=0 ⟹limx→π/2tan(x)tan2x=1

Step-by-step explanation:

Since tanx<0 for π/2<x<π we can assume the function is defined only for 0<x<π/2 (a left neighborhood of π/2).

In all these case it's better to compute the limit of the logarithm:

limx→π2tan2xlogtanx=limx→π2−logcotxcot2x=limx→π2−2cotxlogcotxcot2x−1

by applying tanx=1cotx and the duplication formula

cot2x=cot2x−12cotx.

Now you can substitute t=cotx and get

limt→02tlogt1−t2

and you should know how to manage this form. Hint: the limit is 0.