## Question

###### What is #lim_(x->pi//4) (2 - tanx)^(1/(ln(tanx)))#?

## Answers

#lim_(x->pi/4) (2-tanx)^(1/ln(tanx)) = ?# Well...

#tan (pi/4) = 1# ...so

#lim_(x->pi/4) (2-tanx)^(1/ln(tanx))# is of the form#1^(1/0)# . At this point I just want to move the exponent down. The best way I know is natural logs.

#lim_(x->pi/4) ln[(2-tanx)^(1/ln(tanx))] = ln(2-tanx)/(ln(tanx))# Oh hey, it worked!

#lim_(x->pi/4) ln(2-tanx)/(ln(tanx)) = ?# This is now of the form 0/0, so now we can use

L'Hopital's rule.

#lim_(x->pi/4) ln(2-tanx)/(ln(tanx)) = lim_(x->pi/4) (d/(dx)[ln(2-tanx)])/(d/(dx)[ln(tanx)])#

#= lim_(x->pi/4) (1/(2-tanx)*-cancel(sec^2(x)))/(1/(tanx)*cancel(sec^2(x)))#

#= lim_(x->pi/4) ((-1)/(2-tanx))/(1/(tanx))# We're almost there.

#= lim_(x->pi/4) ((-tanx)/(2-tanx)) = (-1)/(2-1) = -1# Then, we can undo the natural log:

#e^[lim_(x->pi/4) ((-tanx)/(2-tanx))] = e^[(-1)/(2-1)] = e^(-1) = 1/e# And that's it!