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!
 A: Substitute $x=1/t$; then, with a simple massage, the limit becomes
$$
\lim_{t\to0^+}
  \frac{\sqrt{1+t}+\sqrt{1+2t}-2\sqrt{1+3t}}
       {\sqrt{1+2t}+1-2\sqrt{1+3t}}
=\lim_{t\to0^+}
  \frac{(1+\frac{1}{2}t)+(1+t)-2(1+\frac{3}{2}t)+o(t)}
       {(1+t)+1-2(1+\frac{3}{2}t)+o(t)}=
\frac{-3/2}{-2}=\frac{3}{4}
$$
using the first order Taylor expansion.
A: HINT:
$$\sqrt{4x+12}=2\sqrt{x+3}$$
Now set $1/x=h\implies h\to0^+$ 
$\sqrt{x+a}-\sqrt{x+b}$
$=\dfrac{\sqrt{1+ah}-\sqrt{1+bh}}{\sqrt h}$
$=\dfrac{1+ah-(1+bh)}{\sqrt h(\sqrt{1+ah}+\sqrt{1+bh})}$
$$\implies\dfrac{\sqrt{x+1}-\sqrt{x+3}+\sqrt{x+2}-\sqrt{x+3}}{\sqrt x-\sqrt{x+3}+\sqrt{x+2}-\sqrt{x+3}}=\dfrac{\dfrac{1-3}{\sqrt{1+h}+\sqrt{1+3h}}+\dfrac{2-3}{\sqrt{1+2h}+\sqrt{1+3h}}}{\dfrac{0-3}{\sqrt{1}+\sqrt{1+3h}}+\dfrac{2-3}{\sqrt{1+2h}+\sqrt{1+3h}}}$$
A: 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 $\frac{479}{640}\approx 0.748438$.
A: Note that $\sqrt{4x + 12} = 2\sqrt{x + 3}$.  This unlocks all of the "secrets."
A: HINT: Divide the numerator and denominator by $\sqrt x$. And use the fact that $x \to \infty$ implies $\frac{1}{ x} \to 0$.
