I've got this problem:
Let $f:\mathbb{R}\rightarrow \mathbb{R}$ so that $ax^2 - 5x^4 \leq f(x) \leq ax^2$, $a \neq 0$
1) Find $\lim_{x \to 0}f(x)$
2) Find $a$ so that $\lim_{x \to 0}\frac{f(x)}{sin^2(x)} = 3$
1) By the squeeze theorem, I've got: $\lim_{x \to 0}ax^2 - 5x^4 = 0$ and $lim_{x \to 0}ax^2 = 0$, so $lim_{x\to0}f(x)=0$
2) Here I'm not sure if I'm doing things right. I've tried by using the squeeze theorem again in this way: $\frac{ax^2 - 5x^4}{sin^2(x)} \leq \frac{f(x)}{sin^2(x)} \leq \frac{ax^2}{sin^2(x)}$
Now I take these limits:
$\lim_{x\to0}\frac{ax^2}{sin^2(x)} = 3$ and $\lim_{x\to0}\frac{ax^2}{sin^2(x)}=3$ and here is where I'm stuck on since I can't figure out how to find $a$.
Is it ok to do, for example:
$\frac{ax^2}{sin^2(x)}=3 \to ax^2=3sin^2(x) \to a=3\frac{sin^2(x)}{x^2}$, take the limit $\lim_{x\to0}3\frac{sin^2(x)}{x^2} = 3$, so $a=3$
And:
$\frac{ax^2-5x^4}{sin^2(x)}=3 \to ax^2-5x^4=3sin^2(x) \to ax^2=3sin^2(x)+5x^4 \to a=3\frac{sin^2(x)}{x^2} + 5x^2$, then take the limit $\lim_{x\to0}3\frac{sin^2(x)}{x^2} + 5x^2 = 3$, so $a=3$ and then by the squeeze theorem, $\lim_{x\to0}\frac{f(x)}{sin^2(x)} = 3$
Any hint will be appreciated, thanks in advance for your time.