What is the limit as #x->pi/2# of #(1+2cotx)^(tanx)#?

I checked wolframalpha and it exists. So let's figure out what it is.

#y = lim_(x->pi/2)(1+2cotx)^tanx#

Generally when you have complicated exponents like this, it's a pretty good idea to use #ln# for the interesting property that exponents can be moved out of the argument and made into coefficients.

#lny = ln (lim_(x->pi/2) (1+2cotx)^tanx)#

A useful property of limits is that logarithms can be brought into limits.

#lny = lim_(x->pi/2) ln((1+2cotx)^tanx)#

Now we can bring that #tanx# down as a coefficient, and then reciprocate it (#tanx -> 1/(cotx)#).

#lny = lim_(x->pi/2) tanxln(1+2cotx)#

#lny = lim_(x->pi/2) ln(1+2cotx)/(cotx)#

Now, using L'Hopital's rule, we can determine the limit by differentiating the numerator and denominator independently.

#lny = lim_(x->pi/2) [1/(1 + 2cotx)*cancel(-)2cancel(csc^2x)]/(cancel(-csc^2x))#

#lny = lim_(x->pi/2) 2/(1 + 2cotx)#

At this point we can solve this by plugging in regular numbers.

#cotx = cosx/sinx#

Plug in #pi/2# to get:

#cot(pi/2) = (cos(pi/2))/(sin(pi/2)) = 0/1 = 0#

Hence:

#lny = lim_(x->pi/2) 2 = 2#

But remember, we modified the original function using #ln#. To finish this up, simply undo the natural logarithm using exponentiation via the operation #e^u# where #u = 2#.

#color(blue)(y = lim_(x->pi/2)(1+2cotx)^tanx = e^2)#