Limit of a function in which square roots are involved $$\lim_{x \to 0}\frac{x+2-\sqrt{2x+4}}{3x-1+\sqrt{x+1}}$$
Could someone please help me solve this problem. 
I tried multiplying by a unity factor but I end up stuck. 
 A: You may use $$\sqrt{1+x}=1+\frac x2+O(x^2)$$
First,
$$\sqrt{2x+4}=2\sqrt{1+\frac x2}=2(1+\frac x4)+O(x^2)=2+\frac x2+O(x^2)$$
And
$$\frac{x+2-\sqrt{2x+4}}{3x-1+\sqrt{x+1}}
=\frac{x+2-2-\frac x2+O(x^2)}{3x-1+1+\frac x2+O(x^2)}
\\=\frac{\frac x2+O(x^2)}{\frac72x+O(x^2)}=\frac17+O(x)\underset{x\to0}\longrightarrow\frac17
$$
A: $$\lim_{x \to 0}\frac{x+2-\sqrt{2x+4}}{3x-1+\sqrt{x+1}}=\lim_{x\rightarrow0}\frac{((x+2)^2-(2x+4))(3x-1-\sqrt{x+1})}{((3x-1)^2-(x+1))(x+2+\sqrt{2x+4})}=$$
$$=\lim_{x\rightarrow0}\frac{(x^2+2x)(3x-1-\sqrt{x+1})}{(9x^2-7x)(x+2+\sqrt{2x+4})}=\lim_{x\rightarrow0}\frac{(x+2)(3x-1-\sqrt{x+1})}{(9x-7)(x+2+\sqrt{2x+4})}=\frac{2\cdot(-2)}{-7\cdot4}=\frac{1}{7}.$$
A: You can simplify it:
$$\lim_{x \to 0}\frac{x+2-\sqrt{2x+4}}{3x-1+\sqrt{x+1}}=\lim_{x \to 0}\frac{x+2-\sqrt{2}\sqrt{x+2}}{3(x+1)+\sqrt{x+1}-4}=\\
\lim_{x \to 0}\frac{\sqrt{x+2}(\sqrt{x+2}-\sqrt{2})}{3(\sqrt{x+1}-1)(\sqrt{x+1}+\frac43)}=\\
\lim_{x \to 0}\frac{\sqrt{x+2}-\sqrt{2}}{\sqrt{x+1}-1}\cdot \lim_\limits{x\to 0}\frac{\sqrt{x+2}}{3\sqrt{x+1}+4}=\\
\frac{\sqrt{2}}{7}\cdot \lim_{x \to 0}\frac{\color{red}{\sqrt{x+2}-\sqrt{2}}}{\color{blue}{\sqrt{x+1}-1}}=\cdots$$
Can you continue?
Answer:

 Multiply both top and bottom by conjugates:$$\frac{\sqrt{2}}{7}\cdot\lim_{x \to 0}\frac{\require{cancel}\cancel{(\color{red}{(x+2)-2})}(\sqrt{x+1}+1)}{\cancel{(\color{blue}{(x+1)-1})}(\sqrt{x+2}+\sqrt{2})}=\frac{\sqrt{2}}{7}\cdot \frac1{\sqrt{2}}=\frac17.$$

