$\lim_{x\to 0} \frac{\tan ^nx - \sin ^m x}{x^r}=?$without l'Hôpital's rule.

Question

I need to solve without l'Hôpital's rule.

$$n,m,r \in \mathbb{N}$$

$$\lim_{x\to 0} \frac{\tan ^nx - \sin ^m x}{x^r}=?$$

MY TRY :

$$\lim_{x\to 0} \frac{\tan ^nx - \sin ^m x}{x^r}=\\\lim_{x\to 0}\frac{\tan^n x}{x^r}-\frac{\sin^m}{x^r}=?$$

Answer 1

The simplest case is the one when $m = n > 0$ and $r = m + 2$. Then we have $$\lim_{x \to 0}\frac{\tan^{n}x - \sin^{n} x}{x^{n + 2}} = \lim_{x \to 0}\frac{\tan x - \sin x}{x^{3}}\cdot \sum_{i = 0}^{n - 1}\frac{\tan^{n - 1 - i}x}{x^{n - 1 - i}}\cdot\frac{\sin ^{i}x}{x^{i}} = \frac{1}{2}\cdot\sum_{i = 0}^{n - 1}1 = \frac{n}{2}$$ The first limit is evaluated as $1/2$ without using L'Hospital's Rule.

From the above we can see that if $m = n > 0$ and $r > m + 2$ then the limit diverges to $\pm\infty$ and if $r < m + 2$ then it converges to $0$. Dealing with $m \neq n$ is clumsy and it is better to deal with individual values of $m, n$.

Answer 2

HINT

Use $$\lim_{x \rightarrow 0} \frac {sinx}{x}=1$$

There are several cases to study, for example if $n,m \ge r$ then the limit is $0$. Also, for some $n,m,r$ the limit may not exist.