limits and derivatives

Method 1

You should observe that both the numerator and denominator would be equal to zero when x = 0.

Thus, we can evaluate the result by L' Hospital Rule

lim x->0 (e^(2x)-e^(-x))/(xe^x)
= lim x->0 [2e^(2x) + e^(-x)] / (e^x + xe^x)
= (2 + 1) / (1 + 0)
= 3


Method 2

lim x->0 (e^(2x)-e^(-x))/(xe^x)
= lim x->0 [e^(2x) - 1/e^x] / (xe^x)
= lim x->0 [(e^(3x) - 1) / e^x] / (xe^x)
= lim x->0 [e^(3x) - 1] / (x e^2x)
= lim x->0 [(e^x - 1)(e^2x + e^x + 1)] / (x e^2x)
= lim x->0 [(e^x - 1) / (x e^2x)] * (e^2x + e^x + 1)
= lim x->0 [(e^x - 1) / (x e^2x)] * lim x->0 (e^2x + e^x + 1)
= lim x->0 [(e^x) / (e^2x + 2 x e^2x)] * lim x->0 (e^2x + e^x + 1)
= 1 / (1 + 0) * (1 + 1 + 1)
= 3

[e^(2x) - e^(-x)]/xe^x
= [e^2x - 1/e^x]/xe^x
= [e^3x - 1]/xe^2x
= (e^x - 1)(e^2x + e^x + 1)/(xe^2x)
= [(e^x - 1)/x][(e^2x + e^x + 1)/e^2x]
= [(e^x - 1)/x][1 + 1/e^x + 1/e^2x]
so when x tends to zero, the expression = (1)(1 + 1 + 1) = 3.

Using L'Hopital rule
lim x->0 [e^(2x)-e^(-x)]/(xe^x)
=lim x->0 [2e^(2x)+e^(-x)]/(xe^x+e^x)
=3