LIM (x tends to 1) (E-E ^ x) / (x-1)=

LIM (x tends to 1) (E-E ^ x) / (x-1)=

Using the law of Robita:
(e-e^x)'/(x-1)'=(-e^x)/1=-e^x
Let x = 1, - e ^ x = - E
lim[(e -e^x)/(x-1)]=-e

Lim △ x tends to 0 {[e ^ (x + △ x)] - e ^ x} / △ X

lim(△x->0){[e^(x+△x)-e^x]/△x}
=lim(△x->0)[(e^x*e^△x-e^x)/△x]
=LIM (△ X - > 0) {[e ^ x (1 + △ X / 1! + △ X & # / 2! + △ X & # / 3! +) - e ^ x] / △ x} (expand e ^ △ x into series)
=lim(△x->0)[e^x(1/1!+△x/2!+△x²/3!+.)]
=e^x(1/1!+0/2!+0²/3!+.)
=e^x.
Note: this problem is the derivation of the derivative of e ^ x with the definition of derivative

LIM (x tends to 0) 1-x ^ 2-e ^ (- x ^ 2) / xsin ^ 3 (2x)

lim(x→0) [1-x^2-e^(-x^2)]/[xsin^3(2x)] =lim(x→0) [1-x^2-e^(-x^2)]/(8x^4) (0/0)=lim(x→0) [-2x+2xe^(-x^2)]/(32x^3) =lim(x→0) [-1+e^(-x^2)]/(16x^2) =lim(x→0) (-x^2)/(16x^2) =-1/16