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, $\sin\frac{x+\pi}{x-\pi}\sin\frac{x-\pi}{x+\pi} = 0$.

Answer 1:
Set $x=-2+y$
$$\lim_{x\to -2} \frac{\sqrt{6+x}-2}{\sqrt{3+x}-1} = \lim_{y\to 0} \frac{\sqrt{4+y}-2}{\sqrt{1+y}-1}$$
$$=\lim_{y\to 0} \frac{(y/4)+O(y^2)}{(y/2)+O(y^2)}=\lim_{y\to 0} \frac{y/4}{y/2}=(1/2)$$
Answer 2:
Set $x=\pi+y$
$$\lim_{x\to \pi} \sin\frac{x+\pi}{x-\pi}\sin\frac{x-\pi}{x+\pi} = \lim_{y\to 0} \sin\frac{y+2\pi}{y}\sin\frac{y}{y+2\pi} =I $$
$$|I|=\lim_{y\to 0} \left|\sin\frac{y+2\pi}{y}\right|\left|\sin\frac{y}{y+2\pi}\right|\le \lim_{y\to 0} 1\left|\sin\frac{y}{2\pi}\right|=0$$
So $I=0$
