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$$\lim_{x\to\infty}\sqrt{(x^2 + ax)} - \sqrt{(x^2 - ax)} = 4$$

Problem: Solve for a.

I'm taking AP Calculus next year and I wanted to get a bit ahead over the summer, but I got stuck on this problem. I really don't know how to go about this problem, so all I've done (which is probably wrong) is rewrite it so it looks like $\displaystyle \lim_{x\to\infty} (x^2+ax)^{\frac12} - \lim_{x\to\infty}(x^2 - ax)^{\frac12} = 4$ and try to solve for a by squaring both sides and treating it as a binomial. Was I going in the right direction? What should I do?

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    $\begingroup$ Multiply by $1$ or... $$\frac{\sqrt{x^2+ax}+\sqrt{x^2-ax}}{\sqrt{x^2+ax}+\sqrt{x^2-ax}}.$$ $\endgroup$ – Brad Jul 8 '14 at 18:01
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    $\begingroup$ For a second comment, what you did is wrong. You split your first limit which does exist into two limits that are infinite. You are trying to evaluate the limit by $\infty - \infty$ which is called an indeterminate form. $\endgroup$ – Brad Jul 8 '14 at 18:04
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Answer without details

We multiply by the conjugate and using that $$\sqrt{x^2\pm ax}\sim_\infty x$$ then the limit becomes

$$\lim_{x\to\infty}\frac{2ax}{2x}=\boxed{a=4}$$

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  • $\begingroup$ Can you be a bit more specific? What did you multiply by the conjugate ? How did you get to the limit at the bottom? $\endgroup$ – bandicoot12 Jul 8 '14 at 17:59
  • $\begingroup$ We multiply the given expression by this expression $\sqrt{x^2+ax}+\sqrt{x^2-ax}$ and then we divide by it. $\endgroup$ – user63181 Jul 8 '14 at 18:04
  • $\begingroup$ Isn't it supposed to be $\sqrt{x^2\pm ax} \sim_\infty x \pm a/2$? Not that it changes the final answer. $\endgroup$ – Brad Jul 8 '14 at 18:06
  • $\begingroup$ Sweet and simple, @Sami! $\endgroup$ – Namaste Jul 10 '14 at 12:09
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Well you can do it that way also $$x^2+ax-2\sqrt{x^4-a^2x^2}+x^2-ax=16\\x^2-\sqrt{x^4-a^2x^2}=8\\x^2-8=\sqrt{x^4-a^2x^2}\\x^4-16x^2+64=x^4-a^2x^2\\64=(a^2-16)x^2$$ Since $x\to \infty$ means that $a^2-16=0$ otherwise it can't be equal to a finite number,now if $a=-4$ clearly the first square root is smaller than second so $a=4$

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HINT As the others have mentioned multiply by the conjugate which is

$$\frac{\sqrt{x^2+ax}+\sqrt{x^2-ax}}{\sqrt{x^2+ax}+\sqrt{x^2-ax}}$$

so you will have something like:

$$\lim_{x\to\infty}\left(\sqrt{(x^2 + ax)} - \sqrt{(x^2 - ax)}\right) \cdot \frac{\sqrt{x^2+ax}+\sqrt{x^2-ax}}{\sqrt{x^2+ax}+\sqrt{x^2-ax}} = 4$$

The reason you want to do this is so that you can get rid of the pesky square roots which simplifies the question a lot. Try it.

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