how do you solve lim ( arcsin (3x) / tan (5x)) as x tends to 0 using substitution?
First, a little work:
$\displaystyle \lim_{x\to0}\frac{\arcsin3x}{\tan5x}=\lim_{x\to0}\ frac{\arcsin3x\cos5x}{\sin5x}=\left(\lim_{x\to0}\f rac{\arcsin3x}{\sin5x}\right)\cdot\left(\lim_{x\to 0}\cos5x\right)=\lim_{x\to0}\frac{\arcsin3x}{\sin5 x}$
Now let $\displaystyle x=\frac{\sin y}{3}$. It's clear that $\displaystyle x\to0$ implies $\displaystyle y\to0$.
We have now:
$\displaystyle \lim_{x\to0}\frac{y}{\sin\left(\frac{5}{3}\sin y\right)}$
Can you take it from here?
Hint: Multiply and divide by $\displaystyle \frac{5}{3}\sin y$ and regroup the terms to something more amenable.
Or exactly the same as in your other post
$\displaystyle \lim_{x\to0}\frac{\arcsin(3x)}{\tan(5x)}=\lim_{x\t o0}\frac{\tfrac{\arcsin(2x)-\arcsin(2\cdot0)}{x}}{\tfrac{\tan(5x)-\tan(5\cdot0)}{x}}=\frac{\bigg[\arcsin(2x)\bigg]'\bigg|_{x=0}}{\bigg[\tan(5x)\bigg]'\bigg|_{x=0}}$
You really need to try to to take some of the methods posted in your other quetsions and apply them, instead of mass posting every limit you see.