Limit of $\frac{\arcsin{3x}}{\tan{5x}}$ as $x$ goes to zero Can $\displaystyle \lim_{x \to 0}\frac{\arcsin{3x}}{\tan{5x}}$ be done by just pure manipulation? I have done it with L'hopital:
$\displaystyle \lim_{x \to 0}\frac{\arcsin{3x}}{\tan{5x}} = \lim_{x \to 0}\frac{3\cos^2{x}}{5\sqrt{1-9x^2}}=\frac{3\cos^2(0)}{5\sqrt{1-0}} = \frac{3}{5}.$
And in my notes it's done with series expansion. But I feel there should be a way to do this without either of these methods (by more elementary methods). 
 A: Note that $\displaystyle\lim_{x \to 0}\dfrac{\arcsin 3x}{\tan 5x} = \lim_{x \to 0}\dfrac{3}{5} \cdot \dfrac{\arcsin 3x}{3x} \cdot \dfrac{5x}{\tan 5x} = \dfrac{3}{5}\left[\lim_{x \to 0}\dfrac{\arcsin 3x}{3x}\right]\left[\lim_{x \to 0}\dfrac{5x}{\tan 5x}\right]$, provided of course that both of these limits exist. 
The first of these limits can be handled by substituting $x = \dfrac{1}{3}\sin y$, which gives us $\displaystyle\lim_{x \to 0}\dfrac{\arcsin 3x}{3x} = \lim_{y \to 0}\dfrac{y}{\sin y} = 1$. 
The second limit can be handled by substituting $x = 5z$, which gives us $\displaystyle\lim_{x \to 0}\dfrac{5x}{\tan 5x} = \lim_{z \to 0}\dfrac{z}{\tan z} = \lim_{z \to 0}\dfrac{z\cos z}{\sin z} = \left[\lim_{z \to 0}\dfrac{z}{\sin z}\right]\left[\lim_{z \to 0}\cos z\right] = 1 \cdot 1 = 1$. 
Therefore, $\displaystyle\lim_{x \to 0}\dfrac{\arcsin 3x}{\tan 5x} = \dfrac{3}{5}$.
A: The shortest is with equivalents, if you've seen Taylor's formula: $\arcsin u \sim_0 u \sim_0\tan u$, hence
$$\frac{\arcsin 3x}{\tan 5x}\sim_0\frac{3x}{5x}=\frac 35.$$
