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Why does the outcome of the limit as x approaches infinity of $$\sqrt{x^2+2x}- \sqrt{x^2-2x}$$

which simplifies to

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

change depending on whether we take the limit from minus infinity or positive infinity?

If positive, the answer is 2. If negative infinity, the answer is -2??

It seems to me that we get $$ \frac {4}{\sqrt{1} + \sqrt{1}}$$ no matter which side we approach.

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3 Answers 3

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No, you're missing an absolute value sign on $x$ in the denominator.

Put differently, note that for large values of $x$,

$$\sqrt{x^2+2x} \approx |x| + \text{sgn}(x)$$

and

$$\sqrt{x^2-2x} \approx |x| - \text{sgn}(x).$$

Take the difference and you get your limit.

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The equation $$\sqrt{x^2+2x}=x\sqrt{1+\frac{2}{x}}$$ is true only for $x>0$. The correct equation is $$\sqrt{x^2+2x}=\sqrt{x^2}\sqrt{1+\frac{2}{x}}=|x|\sqrt{1+\frac{2}{x}}$$

The intuition is that for $x>0$ you are subtracting a smaller number from a larger one (i.e. the two square roots); however for $x<0$ you are subtracting a larger number from a smaller one.

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  • $\begingroup$ So all of this thread is nonsense? math.stackexchange.com/questions/652245/… Because x is approaches MINUS infinity and so is definitely not greater than 0. $\endgroup$
    – Paze
    Feb 1, 2014 at 18:21
  • $\begingroup$ I mean this math.stackexchange.com/questions/658576/… $\endgroup$
    – Paze
    Feb 1, 2014 at 18:31
  • $\begingroup$ The thread is not nonsense, since $x^2+2x>0$ whether $x\to \infty$ or $x\to -\infty$. $\endgroup$
    – vadim123
    Feb 1, 2014 at 18:32
  • $\begingroup$ Vadim, I am having problems understanding why $\sqrt{x^2}$ plays a part in $$\dfrac {4}{\sqrt{1 + \frac 2x} + \sqrt {1 - \frac 2x}}$$ Which is my end result. I don't see $\sqrt{x^2}$ anywhere here. $\endgroup$
    – Paze
    Feb 1, 2014 at 18:35
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    $\begingroup$ Your end result goes to 4 for $x\to\pm \infty$, but your end result does not equal the starting function for $x<0$. $\endgroup$
    – vadim123
    Feb 1, 2014 at 18:49
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$$\sqrt{x^2} = |x|$$ So when $x$ tends to $- \infty$, $\sqrt{x^2} = -x$. Hence then answer would be: $$\frac{4}{-2} = -2$$

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  • $\begingroup$ Once I have simplified, there is no $\sqrt{x^2}$ why are you bringing that up? It's gone. I'm misunderstanding something. In the simplified version it seems that $\frac{2}{x}$ tends to 0 and therefore we should get 2. $\endgroup$
    – Paze
    Feb 1, 2014 at 18:25
  • $\begingroup$ But how did you simplify that $\sqrt{x^2} = x$? So your initial step itself is wrong. $\endgroup$
    – lsp
    Feb 1, 2014 at 18:31
  • $\begingroup$ Well let's say that I simplify it to $|x|$. Can I still cancel it out like I did or...? I don't even know that with absolute value.. $\endgroup$
    – Paze
    Feb 1, 2014 at 18:53

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