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?

  • 1
    $\begingroup$ Rationalize by multiplying $$\frac{(\sqrt{x^3+1} + \sqrt{4x+1})(\sqrt{x^3-2x}+\sqrt{x+2})}{(\sqrt{x^3+1} + \sqrt{4x+1})(\sqrt{x^3-2x}+\sqrt{x+2})}.$$ $\endgroup$ – Hw Chu Sep 10 '19 at 1:06
  • $\begingroup$ Thank you! I found the solution. If I have time, I will answer my own question. $\endgroup$ – Yip Jung Hon Sep 10 '19 at 1:24
  • $\begingroup$ Its a 0/0 form, so I guess we are good to apply L'Hopital rule as well. $\endgroup$ – Siddhant Sep 10 '19 at 4:12
  • $\begingroup$ Well technically they haven't taught L'Hopital's Rule, and I can imagine it can get very messy. Anyways we are encouraged not to use L'Hopital's Rule $\endgroup$ – Yip Jung Hon Sep 10 '19 at 7:11

$${{\sqrt{x^3+1}-\sqrt{4x+1} \over \sqrt{x^3-2x} - \sqrt{x+2}} = \left({\sqrt{x^3+1}-\sqrt{4x+1} \over \sqrt{x^3-2x} - \sqrt{x+2}} \right) \left( {\sqrt{x^3+1}+\sqrt{4x+1} \over \sqrt{x^3+1}+\sqrt{4x+1}} \right) \left({\sqrt{x^3-2x} + \sqrt{x+2} \over \sqrt{x^3-2x} + \sqrt{x+2}} \right) =\left({x^3-4x \over x^3-3x-2}\right) \left({\sqrt{x^3-2x}+\sqrt{x+2} \over \sqrt{x^3+1}+\sqrt{4x+1}} \right) =\left({x(x+2) \over (x+1)^2}\right) \left({\sqrt{x^3-2x}+\sqrt{x+2} \over \sqrt{x^3+1}+\sqrt{4x+1}} \right)}$$

At $x=2$, above simplified to:

$\displaystyle \left({2 \times 4 \over 3 \times 3}\right) \left({2+2 \over 3+3} \right) = \left({8 \over 9}\right) \left({2 \over 3}\right) = {16 \over 27}$

  • $\begingroup$ Nice, today I learnt a new technique of doing limit questions $\endgroup$ – Yip Jung Hon Sep 10 '19 at 1:52

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.

  • $\begingroup$ Long division step might not be necessary. Just divide y numerator and denominator by t. At t=0, y = (-4/3) / (-9/4) = 16/27 $\endgroup$ – albert chan Sep 10 '19 at 12:55
  • $\begingroup$ @albertchan. This is a key point to me. If you have time to waste, have a look at matheducators.stackexchange.com/questions/8339/… $\endgroup$ – Claude Leibovici Sep 10 '19 at 16:18

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