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How do I solve? I've tried to multiply and divide by the conjugate cannot advance. $$\lim_{x\rightarrow +\infty} \sqrt{(x-a)(x-b)}-x$$

2 Answers 2

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Multiplying and dividing by the conjugate works fine. Let $x$ be positive and larger than $a$ and $b$. We quickly obtain $$\frac{-ax-bx+ab}{\sqrt{(x-a)(x-b)}+x}.$$ Divide top and bottom by $x$. (That is another commonly useful kind of move.) We get $$\frac{-a-b+\frac{ab}{x}}{\sqrt{\left(1-\frac{a}{x}\right)\left(1-\frac{b}{x}\right)}+1}.$$ Now finding the limit is straightforward.

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    @ThomasAndrews: Yes, thank you, having view problems on my computer.2012-11-07
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    Could you explain this a bit more? How does the problem have a conjugate? does that need a fraction?2012-11-09
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    I was answering the OP, who knew about the conjugate. Actually, it is not *quite* the conjugate, but I went along with it. Multiply top and bottom (which is $1$) by $\sqrt{(x-a)(x-b)}+x$. On top we get $(x-a)(x-b)-x^2$, which simplifies to $-ax-bx+ab$. Multiplying by a conjugate is a widely useful trick. When a problem involves $a+b\sqrt{d}$, it is often useful to get $a-b\sqrt{d}$ involved. You may have seen this with complex numbers. There, when you see $a+bi$, the number $a-bi$ is often helpful.2012-11-09
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    OH, I was a bit confused on the terms, I know to mulitply by one, but was confused because I thought conjugates where just for fractions.2012-11-09
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$\bf{My\; Solution::}$ Given $\displaystyle \lim_{x\rightarrow \infty}\sqrt{(x-a)(x-b)}-x\;,$ Now Using $\bf{A.M\geq G.M}$

Now When $x\rightarrow \infty,$ Then $(x-a)\;,(x-b)\rightarrow \infty$

So $\displaystyle \frac{(x-a)+(x-b)}{2}\geq \sqrt{(x-a)(x-b)}\Rightarrow x-\left(\frac{a+b}{2}\right)\geq \sqrt{(x-a)(x-b)}$

And equality hold when $(x-a) = (x-b)\rightarrow \infty$

$\displaystyle \Rightarrow \lim_{x\rightarrow \infty}x-\left(\frac{a+b}{2}\right)-x =\lim_{x\rightarrow \infty} \sqrt{(x-a)(x-b)}-x$

So $\displaystyle \lim_{x\rightarrow \infty}\sqrt{(x-a)(x-b)}-x = -\left(\frac{a+b}{2}\right)$