Calculus Limits Problem L'Hopital's Rule is not allowed.

Question 1: 
  $$\lim_{x\to -2} \frac{\sqrt{6+x}-2}{\sqrt{3+x}-1} =  \ ?$$

I tried to cross multiply $\frac{\sqrt{6+x}-2}{\sqrt{3+x}-1}$with $\frac{\sqrt{3+x}+1}{\sqrt{6+x}+2}$ and I got $x+2$ on both LHS and RHS thus I conclude
$\frac{\sqrt{3+x}+1}{\sqrt{6+x}+2}$ is equal to $\frac {\sqrt{6+x}-2)}{\sqrt{3+x}-1}$.
Thus when I sub $x = -2$ into $\frac{\sqrt{3+x}+1}{\sqrt{6+x}+2}$ I got $0.5$.

Question 2:
  $$\lim_{x\to \pi} \sin\frac{x+\pi}{x-\pi}\sin\frac{x-\pi}{x+\pi} = \ ?$$

I said that $\sin\frac{x-\pi}{x+\pi}$ tends to $0$.
For $\sin\frac{x+\pi}{x-\pi}$, I use squeeze theorem and said that it can lies between $1$ and $-1$.
Thus, $\sin\frac{x+\pi}{x-\pi}\sin\frac{x-\pi}{x+\pi} = 0$.

For real $x, \left|\sin\dfrac{x+\pi}{x-\pi}\right|\le1$ 
$$\lim_{x\to\pi}\sin\frac{x-\pi}{x+\pi}\sin\dfrac{x+\pi}{x-\pi}$$
$$=0\cdot\text{ a finite real number}=0$$

$$\lim_{x\to-2}\frac{\sqrt{6+x}-2}{\sqrt{3+x}-1}=\lim_{x\to-2}\frac{\sqrt{3+x}+1}{\sqrt{6+x}+2}\cdot\lim_{x\to-2}\frac{6+x-4}{3+x-1}$$
$$=\frac{\sqrt{3+(-2)}+1}{\sqrt{6+(-2)}+2}\cdot1$$ as $x+2\ne0$ as $x\ne-2$ as $x\to-2$
