# Finding limit without using $(a-b)(a+b)$ method

I'm working on this problem: $$\lim_{x\to \infty} (\sqrt {x^2 + 2x} - \sqrt {x^2 - 4x})$$

I tried the following approach and currently it's wrong: $$\sqrt {x^2 + 2x} - \sqrt {x^2 - 4x}$$

Taking out the $x^2$ $$\sqrt {x^2 (1 + \frac{2}{x})} - \sqrt {x^2 (1 - \frac{4}{x}})$$

Refine $$\sqrt{x^2 } \sqrt {(1 + \frac{2}{x})} - \sqrt{x^2} \sqrt {(1 - \frac{4}{x}})$$

And $\sqrt{x^2} = x$, then take out the $x$ $$x(\sqrt {(1 + \frac{2}{x})} - \sqrt {(1 - \frac{4}{x}}))$$

And now if I take the limit when $x$ approaches infinity, I supposed that I should have gotten the following result: $$\infty (\sqrt{1+0} - \sqrt{1-0}) = \infty (1-1) = \infty \cdot 0 = 0$$

And it's incorrectly according to the answer sheet. What did I do wrong?

• $\infty\cdot0$ can be anything. Mar 20, 2016 at 20:58
• @Vincenzo Oliva Ok, got it. Mar 20, 2016 at 21:00
• You need a better approximation. $x^2+2x$ is almost $(x+1)^2$ and $x^2-4x$ is almost $(x-2)^2$, so the answer looks likely to be 3. You just need to formalise that ... Mar 20, 2016 at 21:00
• @almagest It is. That was a quick and nice approach. Thanks! Mar 20, 2016 at 21:07

You are not justified in saying that $\infty \times 0 = 0$.
For example, $$x \times \frac{1}{x} \to 1$$ as $x \to \infty$.
The wrong is the assumption that $\infty \times 0 = 0$. You are almost there. Put $y = \dfrac{1}{x}$, then $L = \displaystyle \lim_{y \to 0} \dfrac{\sqrt{1+2y}-\sqrt{1-4y}}{y}$. You can find this limit by one time use of L'hospitale rule.
your wrong is this: when $∞×0=0$ that $0$ be absolute but in your example it tends to $0$. for solving this kind of limit you can use the square formula: $$\sqrt {{x}^{2}+2\,x}-\sqrt {{x}^{2}-4\,x}=(x+1)-(x-2)=3.$$
To finish (using your approach), $\sqrt{1+2/x} = 1+ x^{-1} + o(x^{-1})$ as $x\to\infty$ (by Taylor approximation); likewise $\sqrt{1-4/x} = 1-2x^{-1}+o(x^{-1})$ as $x\to\infty$. So $$\sqrt{x^2+2x}-\sqrt{x^2-4x} = x(1+x^{-1}-1+2x^{-2}+o(x^{-1}))=3+o(1)$$ as $x\to\infty$.