Finding $\displaystyle \lim_{x \to 2}\frac{\sqrt{x^3+1}-\sqrt{4x+1}}{\sqrt{x^3-2x}-\sqrt{x+2}}$ I came across this question. 

Evaluate the limit  $$ \lim_{x \to 2}\frac{\sqrt{x^3+1}-\sqrt{4x+1}}{\sqrt{x^3-2x}-\sqrt{x+2}}$$

I tried rationalizing the denominator, substitution, yet nothing seems to cancel out with the denominator. I don't think we are supposed to use squeeze theorem or L'Hopital rule for this.
Can someone give me a hint in the right direction?
 A: $${{\sqrt{x^3+1}-\sqrt{4x+1} \over \sqrt{x^3-2x} - \sqrt{x+2}}  
= \left({\sqrt{x^3+1}-\sqrt{4x+1} \over \sqrt{x^3-2x} - \sqrt{x+2}} \right)
\left( {\sqrt{x^3+1}+\sqrt{4x+1} \over \sqrt{x^3+1}+\sqrt{4x+1}} \right) 
\left({\sqrt{x^3-2x} + \sqrt{x+2} \over \sqrt{x^3-2x} + \sqrt{x+2}} \right) 
=\left({x^3-4x \over x^3-3x-2}\right)
\left({\sqrt{x^3-2x}+\sqrt{x+2} \over \sqrt{x^3+1}+\sqrt{4x+1}} \right)  
=\left({x(x+2) \over (x+1)^2}\right)
\left({\sqrt{x^3-2x}+\sqrt{x+2} \over \sqrt{x^3+1}+\sqrt{4x+1}} \right)}$$ 
At $x=2$, above simplified to:
$\displaystyle \left({2 \times 4 \over 3 \times 3}\right) 
\left({2+2 \over 3+3} \right) = \left({8 \over 9}\right) \left({2 \over 3}\right) = {16 \over 27}$ 
A: Using a small trick that I enjoy.
Let $x=t+2$ to make
$$y=\frac{\sqrt{x^3+1}-\sqrt{4x+1}}{\sqrt{x^3-2x}-\sqrt{x+2}}=\frac{\sqrt{4 t+9}-\sqrt{t^3+6 t^2+12 t+9}}{\sqrt{t+4}-\sqrt{t^3+6 t^2+10 t+4}}$$ and now use the binomial expansion or Taylor series around $t=0$.
We have
$$\sqrt{4 t+9}=3+\frac{2 t}{3}-\frac{2 t^2}{27}+O\left(t^3\right)$$
$$\sqrt{t^3+6 t^2+12 t+9}=3+2 t+\frac{t^2}{3}+O\left(t^3\right)$$
$$\sqrt{t+4}=2+\frac{t}{4}-\frac{t^2}{64}+O\left(t^3\right)$$
$$\sqrt{t^3+6 t^2+10 t+4}=2+\frac{5 t}{2}-\frac{t^2}{16}+O\left(t^3\right)$$
So
$$y=\frac{-\frac{4 t}{3}-\frac{11 t^2}{27}+O\left(t^3\right) } {-\frac{9 t}{4}+\frac{3 t^2}{64}+O\left(t^3\right) }$$ Now, using the long division
$$y=\frac{16}{27}+\frac{47 }{243}t+O\left(t^2\right)$$ which shows the limit and also how it is approached.
