Alright, let's evaluate the two given limits.
For the first limit, we want to find lim (x→0) of (1+x-e^x)/(1-cos(x)).
The Taylor series expand of e^x is 1 + x + x^2/2! + x^3/3!....and of cos(x) is 1 - x^2/2! + x^4/4!...
So expanding the limit expression:
(1+x-e^x)/(1-cos(x)) = (1+x-(1 + x + x^2/2! + x^3/3! +...))/(1-(1 - x^2/2! + x^4/4!...))
This simplifies to: (x^2/2! + x^3/3! +...)/(x^2/2! - x^4/4!...)
Dividing every term by x^2 gives: (1/2! + x/3! +...)/(1/2! - x^2/4!...)
Taking the limit as x approaches 0, all terms containing x vanish, leaving only the constant terms:
lim (x→0) = 1/2!
Therefore, lim (x→0) of (1+x-e^x)/(1-cos(x)) is equal to 2.
Next, for the second limit we find lim (x→0) of (x^3)/(tan(x)-x).
The Taylor series expand of tan(x) is x + x^3/3 + 2*x^5/15...
Substituting for tan(x) in the limit expression gives:
lim (x→0) = x^3/(x + x^3/3 + 2*x^5/15 -x)
Simplifying: lim (x→0) = x^3/(x^3/3 + 2*x^5/15)
Dividing every term by x^3 gives: lim (x→0) = 1/(1/3 + 2*x^2/15)
Taking the limit as x approaches 0, all terms containing x vanish, leaving only the constant term:
lim (x→0) = 1/(1/3)
Therefore, lim (x→0) of (x^3)/(tan(x)-x) is equal to 3.