Limit of $\frac{\sin^2(x^p)}{x^q+x^r}$ as $x \to 0^+$ 
Suppose $p,q,r \in \mathbb{R}$ with $p > 0$. Write (without proof) the limit of the following expression: $$\lim_{x \to 0^+} \frac{\sin^2(x^p)}{x^q+x^r}$$ In particular, find and prove the limit of the expression if $p=3$, $q=6$, $r=7$.

I got (through countless graphing on Desmos):
$$\lim_{x \to 0^+} \frac{\sin^2(x^p)}{x^q+x^r}=\begin{cases} 0 & \text{if either $q < 2p$ or $r < 2p$} \\ \frac 12 & \text{if $q=r=2p$} \\ 1 & \text{if either $q=2p$, $r > 2p$ or $q > 2p$, $r=2p$} \\ \infty & \text{if $q > 2p$, $r > 2p$} \end{cases}$$
Also, in the case of $p=3$, $q=6$, $r=7$, the expression becomes $$\lim_{x \to 0^+} \frac{\sin^2(x^3)}{x^6+x^7}.$$
I was able to prove that $$\lim_{x \to 0^+} \frac{\sin^2(x^3)}{x^6+x^7} = \lim_{x \to 0^+} \frac{(\sin(x^3))^2}{x^6(1+x)} \le \lim_{x \to 0^+} \frac{(x^3)^2}{x^6(1+x)}=\lim_{x \to 0^+}\frac 1{1+x} = 1.$$ Now, the graph of this function on Desmos tells me that, in fact, $$\lim_{x \to 0^+} \frac{\sin^2(x^3)}{x^6+x^7} = 1.$$ How can I prove the other inequality, to achieve this result? Namely, I have trouble with proving $$\lim_{x \to 0^+} \frac{\sin^2(x^3)}{x^6+x^7} \ge 1.$$

It is well known that $$\frac{\sin\left(x\right)}{x}\longrightarrow1
 $$ at $x\rightarrow0
 $, then $$\lim_{x\rightarrow0^{+}}\frac{\sin^{2}\left(x^{3}\right)}{x^{6}+x^{7}}=\lim_{x\rightarrow0^{+}}\frac{x^{6}}{x^{6}+x^{7}}=1
 $$ no need to split in two inequalities.
