Question

Lim x>0 (x(e^3x - 1)/(2 - 2cosx))

Answer 1

Evaluating the limand directly at x = 0 yields the indeterminate form 0/0. If L'Hopital's rule is known to you, you can compute the limit by applying it twice:

`\displaystyle\lim_(x\to0)(x\left(e^(3x)-1\right))/(2-2\cos(x)) = \lim_(x\to0)(3xe^(3x)+e^(3x)-1)/(2\sin(x)) \\\\\\ = \lim_(x\to0)(9xe^(3x)+6e^(3x))/(2\cos(x)) = \frac62=\boxed{3}`