Without l'Hôpital or series, $\lim_{y\to 0}\big(\sin(\tan y) - \tan y\big)/(my^n)$ $=1\,$?

Question

Without l'Hôpital or series.

I have tried to give a known form, but I cannot manipulate $m$ and $n$ for what is requested.

Any ideas?

$$\lim_{y\to 0}\frac{\sin(\tan(y)) - \tan(y)}{my^n}=1$$

thanks first of all.

Answer 1

If your questions was to find the possible values of $m$ and $n$, then you can do the following: $$\lim_{y\to 0}\frac{\sin(\tan(y)) - \tan(y)}{my^n}=1\\ \implies\lim_{y\to 0}\frac{\sin(\tan(y)) - \tan(y)}{\tan(y)}\frac{\tan(y)}{my^n}=1\\\implies\lim_{n\to0}\frac{1}{my^{n-1}}=-1$$

This gives us that $m=-1$ and $n=1$, considering that $m,n$ are constants.