Find $\lim_{x\rightarrow 0}{\frac{\tan(x)-x}{x\cdot(e^{x^2}-e^{x^3})}}$

Question

I have to solve the following limit:

$$\lim_{x\rightarrow 0}{\frac{\tan(x)-x}{x\cdot(e^{x^2}-e^{x^3})}}$$

I'm struggling to find a solution. I tried using L'Hôpital's rule but it's not working.

I also tried to do some change in the function, for instance splitting it in two addends, but without success.

Answer 1

Thank you all, I upvoted both the answers. Anyway I think the most straightforward solution to me is combining this one with the second comment.

$$\begin{align}L&=\lim_{x\rightarrow 0}{\frac{\tan(x)-x}{x^3}}\cdot\frac{x^2}{(e^{x^2}-e^{x^3})}\\ \\ &=\lim_{x\rightarrow 0}{\frac{\tan(x)-x}{x^3}}\cdot\lim_{x\rightarrow 0}{\left(\frac{e^{x^2}-e^{x^3}}{x^2}\right)^{-1}}\\ \\ &\overset{\mathrm{H}}{=}\lim_{x\rightarrow 0}{\frac{\tan(x)^2}{3x^2}}\cdot\lim_{x\rightarrow 0}{\left(\frac{e^{x^2}-1}{x^2} - \frac{e^{x^3}-1}{x^2}\right)^{-1}}\\ \\ &=\frac{1}{3}\cdot 1=\frac{1}3\end{align}$$

Answer 2

Near $0$,$$\tan(x)=x+\frac{x^3}3+\cdots,$$and$$x\left(e^{x^2}-e^{x^3}\right)=x^3-x^4+\cdots$$Therefore\begin{align}\lim_{x\to0}\frac{\tan(x)-x}{x\left(e^{x^2}-e^{x^3}\right)}&=\lim_{x\to0}\frac{\frac{x^3}3+\cdots}{x^3-x^4+\cdots}\\&=\lim_{x\to0}\frac{\frac13+\cdots}{1-x+\cdots}\\&=\frac13.\end{align}

Answer 3

$L=\lim_{x\rightarrow 0}{\frac{\tan(x)-x}{x(e^{x^2}-e^{x^3})}}$

$$\begin{align}L&=\lim_{x\rightarrow 0}{\frac{\tan(x)-x}{x^3}}\cdot\frac{x^2}{(e^{x^2}-e^{x^3})}\\ \\ &=\lim_{x\rightarrow 0}{\frac{\tan(x)-x}{x^3}}\cdot\lim_{x\rightarrow 0}\frac{x^2}{(e^{x^2}-e^{x^3})}\\ \\ &=\frac{1}{3}\cdot 1=\frac{1}3\end{align}$$