I think you need a division sign not multiplication.
L'Hopital's Rule twice.
Limits to infinity isolate the highest powered exponents in the numerator and denominator.
What is going on for number 4?
1. Limit [x->1] [(e^x-e)*(x-1)] = e
How to solve ? Derivation (L'Hopital's rule) don't get anything useful - it remains [0/0].
2. Limit [x->0] [Ln[Cos[x]] / x^2] = -1/2
Any idea ?
3. Limit [x->Inf] [(x^3+2x-1)^(1/3) / (x+2)] = 1
How to solve ?
4. [a^b]' = b*Ln[a]*[a]'
Is it right?
Derivating whole function I've got :
[(e^x-e)/(x-1)^1]' = (e^x)*(x-1)^(-1) + (e^x-e)(-(x-1)^(-2)) = (e^x(x-1) - e^x+e ) / (x-1)^2 = e^x / (x-1)
So : lim[x->1, (e^x)/(x-1)] = e/0 => we aren't allowd to devide by zero => deadlock
Have I somewhere got mistake ?