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I'm taking the limit as x approaches infinity from the left (-) of:

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

However I'm not sure how to go about this. I'm at:

$$ \sqrt{ \frac{x^3+4x^2}{x+2x}}- \sqrt \frac{x^3-4x^2}{x-2x} $$

But I'm not sure if it's okay that I say:

"Well, $x^3$ grows exponentially faster than the other factors so the first line gives us $ \sqrt{ \infty} - \sqrt { - \infty } $ so undefined??

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Note that $(\sqrt{a}-\sqrt{b})(\sqrt{a}+\sqrt{b}) = a+b$. –  copper.hat Jan 31 at 16:02
    
@Paze, what does x approaches infinity from the left (-) mean? That $\lim\limits_{x\to-\infty}$? –  Poppy Jan 31 at 16:11

2 Answers 2

up vote 6 down vote accepted

Multiply numerator and denominator by $$\sqrt{x^2+2x}+ \sqrt{x^2-2x}$$ That is, multiply by $1$: $$\dfrac{\sqrt{x^2+2x}- \sqrt{x^2-2x}}{1}\cdot \dfrac{\sqrt{x^2+2x}+ \sqrt{x^2-2x}}{\sqrt{x^2+2x}+ \sqrt{x^2-2x}} = \dfrac{4x}{\sqrt{x^2+2x}+ \sqrt{x^2-2x}}$$

Now, divide numerator and denominator by $x = \sqrt{x^2}$, since $x \to \infty $ implies $x>0$.

ADDED: $$\dfrac{4x}{\sqrt{x^2+2x}+ \sqrt{x^2-2x}} = \dfrac {4x}{\sqrt{x^2\left(1 + \frac 2{x}\right)} + \sqrt{x^2\left(1 - \frac 2x\right)}} = \ldots$$

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Hm, I end up with $ \frac{0}{stuff} $ is that a legitimate way of evaluating the limit to 0? –  Paze Jan 31 at 16:06
    
@amWhy, just a query why $x$ can not be negative? –  lab bhattacharjee Jan 31 at 16:07
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@labbhattacharjee, as $\;x\to\infty\;$ we can safely assume $\;x>0\;$ . BTW, +1 –  DonAntonio Jan 31 at 16:08
    
SOrry not 0, let me calculate further –  Paze Jan 31 at 16:08
    
Paze: No, be careful $(a + b)(a - b) = a^2 - b^2$ which in this case is $(x^2 + 2x) - (x^2 - 2x) = x^2 - x^2 + 2x - x^2 + 2x = 4x$. –  amWhy Jan 31 at 16:08

Clearly, $x$

HINT: Setting $\displaystyle\frac1x=h,$

$$\lim_{x\to\infty}[\sqrt{x^2+2x}-\sqrt{x^2-2x}]=\lim_{h\to0}\frac{\sqrt{1+2h}-\sqrt{1-2h}}{|h|}$$

Now, $$\sqrt{1+2h}-\sqrt{1-2h}=\frac{1+2h-(1-2h)}{\sqrt{1+2h}+\sqrt{1-2h}}=\frac{4h}{\sqrt{1+2h}+\sqrt{1-2h}}$$

as $\displaystyle h\to0\implies h\ne0\implies\lim_{h\to0}\frac hh=1$

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