Limit with log and trig functions using L'Hopitals rule Could anyone tell me if this is correct?
$$ \lim_{x\to 0} \frac{e^{2x}-\pi^{x}}{sin(3x)} = \lim_{x\to 0} \frac{2e^{2x}-\pi^{x} ln(\pi)}{3cos(3x)} = \frac{2e^{2(0)}-\pi^{(0)} ln(\pi)}{3cos(3(0))} = \frac{2\cdot 1-0}{3\cdot1} = \frac{2}{3} $$
I'm getting a bit of a different result on Wolfram|Alpha
If it's incorrect, mind telling me where I went wrong? Thanks!

Your error is here:
$$\pi^0\ln(\pi)\neq0$$
