To show that: $\lim_{x \to 0}\frac{\sin^{-1}(x)}{\tan^{-1}(3x)} = \frac{1}{3}$ using epsilon delta definition of limit
Attempt:
$ |f(x) - \frac{1}{3}| = |\frac{\sin^{-1}(x)}{\tan^{-1}(3x)} - \frac{1}{3}| = |\frac{3\sin^{-1}(x) -\tan^{-1}(3x)}{3\tan^{-1}(3x)}|$
$\frac{3\sin^{-1}(x) -\tan^{-1}(3x)}{3\tan^{-1}(3x)}$ is an even function and hence we can take $x > 0$
For $x > 0$,
$\frac{3\sin^{-1}(x) -\tan^{-1}(3x)}{3\tan^{-1}(3x)} < \frac{\sin^{-1}(3x) -\tan^{-1}(3x)}{3\tan^{-1}(3x)}$
I tried putting $3x = \tan t$ but got stuck.
I am knew to the epsilon delta definition.
Let $$f(x):=\frac{\sin^{-1}x}x\quad\text{and}\quad g(x):=\frac{\tan^{-1}x}x.$$ I will assume you know that $$\lim_{x\to0}f(x)=\lim_{x\to0}g(x)=1,$$ i.e. for all $\alpha>0,$ there exists $\beta(\alpha)>0$ and $\gamma(\alpha)>0$ such that for every real number $x:$ $$\begin{cases}0<|x|<\beta(\alpha)\implies|f(x)-1|<\alpha\\ 0<|x|<\gamma(\alpha)\implies|g(x)-1|<\alpha.\end{cases}$$ For every $x$ such that $0<|x|<\min\left(\beta(\alpha),\frac13\gamma(\alpha)\right),$ we have $|f(x)-1|<\alpha$ and $|g(3x)-1|<\alpha,$ hence $$\begin{align}\left|\frac{\sin^{-1}(x)}{\tan^{-1}(3x)}-\frac13\right|&=\frac13\left|\frac{f(x)}{g(3x)}-1\right|\\&=\frac{|f(x)-g(3x)|}{3|g(3x)|}\\&=\frac{|(f(x)-1)-(g(3x)-1)|}{3|g(3x)-1+1|}\\&<\frac{2\alpha}{3(1-\alpha)}, \end{align}$$ provided $\alpha<1$ (for the final inequality).
There only remains to show that for every $\epsilon>0,$ there exists some $\alpha\in(0,1)$ such that $\frac{2\alpha}{3(1-\alpha)}\le\epsilon.$ The largest solution is $\alpha=\frac{3\epsilon}{2+3\epsilon}.$