$\ \frac{lim}{x \rightarrow 0} (\frac{1}{3x} - \frac{1}{e^{3x} -1})$
I know I am suppose to use L'Hospital but it hasn't worked for me yet. I thought at first it was DNE because I had $\frac{1}{3x^2}$ for my first term. But that and both negative and positive infinity failed.
You will need to combine the fractions first. $$\lambda=\lim_{x\to 0}\left(\frac{1}{3x}-\frac{1}{e^{3x}-1}\right)$$ $$\lambda=\lim_{x\to 0}\left(\frac{e^{3x}-1-3x}{3x(e^{3x}-1)}\right)\stackrel{\mathcal{H^2}}{=}\dfrac{1}{2}$$ Where $\mathcal{H^2}$ denotes the application of L'Hopital's rule twice.
I combined the fractions $$\lim_{x\to 0}\Biggl(\frac{e^{3x}-1-3x}{3x(e^{3x}-1)}\Biggr)$$ Then I applied L'Hospital rule twice $$\Biggl(\frac{\frac{d}{dx}(e^{3x}-1-3x)}{\frac{d}{dx}(3x(e^{3x}-1))}\Biggr)$$ $$\Biggl(\frac{3(e^{3x}-1)}{e^{3x}(9x+3)-3}\Biggr)$$ $$\Biggl(\frac{\frac{d}{dx}(3(e^{3x}-1))}{\frac{d}{dx}(e^{3x}(9x+3)-3)}\Biggr)$$ $$\Biggl(\frac{9e^{3x}}{9e^{3x}(3x+2)}\Biggr)$$ Now I could plug in zero in to x to get $$\frac{9}{18}$$ Thus the limit equals $$\frac{1}{2}$$
$$=\lim_{3x\to0}\dfrac{e^{3x}-1-3x}{(3x)^2}\cdot\dfrac1{\lim_{x\to0}\dfrac{e^{3x}-1}{3x}}$$
If we want to avoid L'Hôpital Rule or Series Expansion,
use Are all limits solvable without L'Hôpital Rule or Series Expansion for the first limit
