Find limit

We have
$$\lim_{x\to 0}\frac{\sin^2x}{x^n}$$ for $n=0,1,2,3,...$
Clearly, for $n=0$ the limit is not indeterminate and will converge to $\sin^20=0$
For all $n\geq1$, the limit is indeterminate of the form $\frac00$, hence L-Hospital's rule is applicable
Applying the rule we get
$$=\lim_{x\to 0}\frac{2\sin x\cos x}{nx^{n-1}}$$
For $n=1$, the limit is simplified to $$=\lim_{x\to 0}\frac{2\sin x\cos x}{1}=0$$
For $n=2$, the limit is simplified to $$=\lim_{x\to 0}\frac{2\sin x\cos x}{2x}=\lim_{x\to 0}\frac{\sin x}{x}\cos x=1\cdot1=1$$
For all odd integers n such that $n>2$, the limit is simplified to $$=\lim_{x\to 0}\frac{\sin 2x}{nx^{n-1}}=\lim_{x\to 0}\frac{2\cos2x}{n(n-1)x^{n-2}}$$ after using L-Hospital's rule twice,
Now it is no longer indeterminate and $n-2$ will always be odd. So its left and right side limits are goverened by the form $$\frac{1}{\to 0^-}\;\;\;or\;\;\;\frac1{\to 0^+}$$
Since, the limits are not equal, so the limit does not exists for the case n being odd and greater than 2.
For all even integers n such that $n>2$, the limit is simplified to $$=\lim_{x\to 0}\frac{\sin 2x}{nx^{n-1}}=\lim_{x\to 0}\frac{2\cos2x}{n(n-1)x^{n-2}}$$ after using L-Hospital's rule twice,
Now it is no longer indeterminate and $n-2$ will always be even. So its left and right side limits are goverened by the form only $$\frac{1}{\to 0^+}=\to +\infty$$