lim[ln(1+x)+ln(1-x)]/(tanx)^2

lim[ln(1+x)+ln(1-x)]/(tanx)^2

X tends to zero, right
that
Original limit
=LIM (x tends to 0) ln (1-x ^ 2) / (TaNx) ^ 2
Ln (1-x ^ 2) is equivalent to - x ^ 2
(TaNx) ^ 2 is equivalent to x ^ 2
therefore
Original limit
=LIM (x tends to 0) - x ^ 2 / x ^ 2
= -1
So the limit value is - 1

lim/x→3.14/2^+ ln(x-3.14/2)/tanx

LIM (x - > π / 2 +) [ln (x - π / 2) / TaNx] = LIM (x - > π / 2 +) {[1 / (x - π / 2)] / sec & sup2; X} (∞ / ∞ type limit, by using Robida's law)
=LIM (x - > π / 2 +) [cos & sup2; X / (x - π / 2)] (limit of type 0 / 0, applying Robida's law)
=lim(x->π/2+)(-2sinx*cosx)
=lim(x->π/2+)[-sin(2x)]
=-sinπ
=0.

How to solve that the x power of E is equal to x

There is no solution to the equation
Obviously, the equation has no solution when x0
Let f (x) = e ^ X-X
Then f '(x) = e ^ X-1 > 0, that is, f (x) is an increasing function
And f (0) = 1, so when x > 0, f (x) > F (0) = 1
This shows that f (x) = 0 is impossible
It's over