Limit $\lim_{x\to \infty} \frac{\sqrt{x+1}+\sqrt{x+2}-\sqrt{4x+12}}{\sqrt{x+2}+\sqrt{x}-\sqrt{4x+12}}$ Using the equivalency, I got the answer equal to $1$, however the answer is given as $\frac34$. 
WolframAlpha also gives $\frac34$.
Could someone tell me that the given answer is wrong or I made a mistake!

Similar to egreg's answer, considering
$$\frac{\sqrt{x+1}+\sqrt{x+2}-\sqrt{4x+12}}{\sqrt{x+2}+\sqrt{x}-\sqrt{4x+12}}$$ extract $\sqrt x$ for each term (and divide both numerator and denominator by  $\sqrt x$); now substitute $t=\frac 1x$ in each term.
So, the expression becomes $$\frac{\sqrt{1+t}+\sqrt{1+2t}-2\sqrt{1+3t}}
       {\sqrt{1+2t}+1-2\sqrt{1+3t}}$$ with $t\to 0$.
Now, using Taylor or generalized binomial theorem $$\sqrt{1+at}=1+\frac{a t}{2}-\frac{a^2 t^2}{8}+O\left(t^3\right)$$ So, the numerator becomes $$\left(1+\frac{t}{2}-\frac{t^2}{8}+O\left(t^3\right)\right)+\left(1+\frac{2t}{2}-\frac{4 t^2}{8}+O\left(t^3\right)\right)-2\left(1+\frac{3 t}{2}-\frac{9 t^2}{8}+O\left(t^3\right)\right)$$ that is to say $$-\frac{3 t}{2}+\frac{13 t^2}{8}+O\left(t^3\right)$$
For the denominator $$\left(1+\frac{2t}{2}-\frac{4 t^2}{8}+O\left(t^3\right)\right)+1-2\left(1+\frac{3 t}{2}-\frac{9 t^2}{8}+O\left(t^3\right)\right)=-2 t+\frac{7 t^2}{4}+O\left(t^3\right)$$ So,$$\frac{\sqrt{1+t}+\sqrt{1+2t}-2\sqrt{1+3t}}
       {\sqrt{1+2t}+1-2\sqrt{1+3t}}=\frac{-\frac{3 t}{2}+\frac{13 t^2}{8}+O\left(t^3\right) }{-2 t+\frac{7 t^2}{4}+O\left(t^3\right) }=\frac{-\frac{3 }{2}+\frac{13 t}{8}+O\left(t^2\right) }{-2 +\frac{7 t}{4}+O\left(t^2\right) }$$ Performing the long division, you will then get $$\frac{\sqrt{1+t}+\sqrt{1+2t}-2\sqrt{1+3t}}
       {\sqrt{1+2t}+1-2\sqrt{1+3t}}=\frac{3}{4}-\frac{5 t}{32}+O\left(t^2\right)$$
$$\frac{\sqrt{x+1}+\sqrt{x+2}-\sqrt{4x+12}}{\sqrt{x+2}+\sqrt{x}-\sqrt{4x+12}}=\frac{3}{4}-\frac{5 }{32x}+O\left(\frac{1}{x^2}\right)$$ 
which shows the limit and also how it is approached.
Using $x=100$, the exact value is $\approx 0.748456$ while the above approximation gives $\frac{479}{640}\approx 0.748438$.
