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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??

José Carlos Santos
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Paze
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2 Answers2

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

amWhy
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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 '14 at 16:06
  • @amWhy, just a query why $x$ can not be negative? – lab bhattacharjee Jan 31 '14 at 16:07
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    @labbhattacharjee, as $;x\to\infty;$ we can safely assume $;x>0;$ . BTW, +1 – DonAntonio Jan 31 '14 at 16:08
  • SOrry not 0, let me calculate further – Paze Jan 31 '14 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 '14 at 16:08
  • @lab: READ my post..."since $x\rightarrow \infty$ implies $x>0."$ – amWhy Jan 31 '14 at 16:09
  • The last part of your solution confuses me. If I divide by $ \sqrt{x^2} $ what is supposed to happen? – Paze Jan 31 '14 at 16:12
  • See how I approached the problem in my "edit/added" portion. All you need now is to factor out $\sqrt{x^2} = x$, which then cancels with the factor of $x$ in the numerator, $$ \dfrac {4x}{x \left(\sqrt{1 + \frac 2x} + \sqrt {1 - \frac 2x}\right) }= \dfrac {4}{\sqrt{1 + \frac 2x} + \sqrt {1 - \frac 2x}}$$ after which you can now take the limit as $x\to \infty$. – amWhy Jan 31 '14 at 16:19
  • Thank you, but the limit is approaching from the left, isn't that worth mentioning? – Paze Jan 31 '14 at 16:23
  • No, because $x$ can only approach $\infty$ from the left! That is, approaching infinity means growing without bound, and to "grow" means moving from "left" to "right". – amWhy Jan 31 '14 at 16:24
  • @amWhy, could it be that Paze thinks $\lim\limits_{x\to-\infty}$? – Poppy Jan 31 '14 at 16:29
  • Because the limit is not the same in $+\infty$ and $-\infty$ in this case. – Poppy Jan 31 '14 at 16:31
0

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

lab bhattacharjee
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