A problem on limit involving various functions Find the value of $$\lim \limits_{x\to 0} \frac{\tan\sqrt[3]x\ln(1+3x)}{(\tan^{-1}\sqrt{x})^2(e^{5\large \sqrt[3]x}-1)}$$
Applying L'Hospital's rule does not seem to simplify the expression.
 A: Note that
$$\lim \limits_{y\to 0}\frac{\tan y}{y}=\lim_{y\to 0}\frac{\tan^{-1} y}{y}=\lim_{y\to 0}\frac{\ln (1+y)}{y}=\lim_{y\to 0}\frac{e^y-1}{y}=1.$$
Therefore,
$$\lim \limits_{x\to 0^+} \frac{\tan\sqrt[3]x\cdot \ln(1+3x)}{\left(\tan^{-1}\sqrt{x}\right)^2\cdot(e^{5 \sqrt[3]x}-1)}=\lim_{x\to 0^+}\frac{\sqrt[3]x\cdot 3x}{(\sqrt{x})^2\cdot 5 \sqrt[3]{x}}=\frac{3}{5}.$$

Edit: Some comments below suggest that the answer above may not be detailed enough. Let me explain a little more about the gap between the two lines of equations above. The limit we are concerned about can be written as
$$\lim \limits_{x\to 0^+}\left(\frac{\tan\sqrt[3]x}{\sqrt[3]x}\cdot \frac{\ln(1+3x)}{3x}\cdot \left(\frac{\sqrt{x}}{\tan^{-1}\sqrt{x}}\right)^2\cdot\frac{5 \sqrt[3]x}{e^{5 \sqrt[3]x}-1}\cdot \frac{\sqrt[3]x\cdot 3x}{(\sqrt{x})^2\cdot 5 \sqrt[3]{x}}\right). $$
On the one hand, as $x\to 0^+$, $\sqrt[3]x$, $3x$ $\sqrt x$ and $5\sqrt[3]x$ all approach to $0$, so from the first displayed line we know that each limit of first four terms in the product above exists and equals $1$. On the other hand, the limit of the last term  exists and equals $\dfrac{3}{5}$. As a result, the original limit exists and equals $\dfrac{3}{5}$.
A: If you change variable (x = y^3) and expand as a Taylor series at y=0, the first terms are   3/5 - 3 y / 2 + 29 y^2 / 20 +...
