I am trying to find $\lim_{x\to 0}\frac{x3^x}{3^x-1}.$
I don't really know where to start, I know I have done this many times but I can never remember what to do for this and there are no practice problems for this in my book.
I am trying to find $\lim_{x\to 0}\frac{x3^x}{3^x-1}.$
I don't really know where to start, I know I have done this many times but I can never remember what to do for this and there are no practice problems for this in my book.
It helps you: $3^x=e^{\ln(3)x}$; (3^x)'=\ln(3)e^{\ln(3)x}=\ln(3)3^x
$ \lim_{x\to0}\frac{x3^x}{3^x-1} = \left(\lim_{x\to0} 3^x \right)\left(\lim_{x\to0}\frac{x}{3^x-1}\right). $ L'Hôpital's rule does the second limit; the first is trivial.
Alternatively, you could say that the second limit is the reciprocal of $\lim\limits_{x\to 0}\dfrac{3^x - 1}{x}$, and that is $\left.\dfrac{d}{dx} 3^x \right|_{x=0}$.