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?
Best Answer
$${{\sqrt{x^3+1}-\sqrt{4x+1} \over \sqrt{x^3-2x} - \sqrt{x+2}} = \left({\sqrt{x^3+1}-\sqrt{4x+1} \over \sqrt{x^3-2x} - \sqrt{x+2}} \right) \left( {\sqrt{x^3+1}+\sqrt{4x+1} \over \sqrt{x^3+1}+\sqrt{4x+1}} \right) \left({\sqrt{x^3-2x} + \sqrt{x+2} \over \sqrt{x^3-2x} + \sqrt{x+2}} \right) =\left({x^3-4x \over x^3-3x-2}\right) \left({\sqrt{x^3-2x}+\sqrt{x+2} \over \sqrt{x^3+1}+\sqrt{4x+1}} \right) =\left({x(x+2) \over (x+1)^2}\right) \left({\sqrt{x^3-2x}+\sqrt{x+2} \over \sqrt{x^3+1}+\sqrt{4x+1}} \right)}$$
At $x=2$, above simplified to:
$\displaystyle \left({2 \times 4 \over 3 \times 3}\right) \left({2+2 \over 3+3} \right) = \left({8 \over 9}\right) \left({2 \over 3}\right) = {16 \over 27}$