Thread: lim x->∞/-∞?

1. lim x->∞/-∞?

Why is it that lim x->∞ $\sqrt {x^2-2x}-x = 1$ while lim x->-∞ $\sqrt {x^2-2x}-x =$∞? They both simplify to the same function but one gives a numerical value while the other one doesn't.

2. Re: lim x->∞/-∞?

Originally Posted by Barthayn
Why is it that lim x->∞ $\sqrt {x^2-2x}-x = 1$ while lim x->-∞ $\sqrt {x^2-2x}-x =$∞? They both simplify to the same function but one gives a numerical value while the other one doesn't.
First a correction: let $f(x)=\sqrt{x^2-2x}-x$ then
$\lim _{x \to \infty } f(x) = - 1$.

Question: Have you graphed the function?

Do you see that $\text{if }x<0\text{ then }f(x)>0~?$

What does the sign of $f'(x)$ when $x<0$ tell you?

3. Re: lim x->∞/-∞?

Hello, Barthayn!

I don't understand your problem.
I get only one answer.

$\text{Why is it that }\lim_{x\to\infty}\left(\sqrt{x^2-2x}-x\right) \:=\: 1$

. . . $\text{while }\lim_{x\to -\infty}\left(\sqrt {x^2-2x} -x\right) \:=\:\infty\,?$

We have: . $\frac{\sqrt{x^2-2x} - x}{1}$

Multiply by $\frac{\sqrt{x^2-2x} + x}{\sqrt{x^2-2x} + x}:\;\;$

. . $\frac{\sqrt{x^2-2x} - x}{1} \cdot\frac{\sqrt{x^2-2x} + x}{\sqrt{x^2-2x} + x} \;=\;\frac{(x^2-2x) - x^2}{\sqrt{x^2-2x} + x} \;=\;\frac{-2x}{\sqrt{x^2-2x} + x}$

Divide numerator and denominator by $x\!:$

. . $\dfrac{\dfrac{-2x}{x}}{\dfrac{\sqrt{x^2-2x}}{x} + \dfrac{x}{x}} \;=\; \dfrac{-2}{\dfrac{\sqrt{x^2-2x}}{\sqrt{x^2}} + 1} \;=\;\dfrac{-2}{\sqrt{\dfrac{x^2-2x}{x^2}} + 1}$

. . $=\;\dfrac{-2}{\sqrt{\dfrac{x^2}{x^2}-\dfrac{2x}{x^2}} + 1} \;=\;\dfrac{-2}{\sqrt{1 - \dfrac{2}{x}} + 1}$

Therefore: . $\lim_{x\to\infty}\dfrac{-2}{\sqrt{1 - \frac{2}{x}} + 1} \;=\;\dfrac{-2}{\sqrt{1+0} + 1} \;=\;\frac{-2}{2} \;=\;-1$

4. Re: lim x->∞/-∞?

Originally Posted by Soroban
Hello, Barthayn!

I don't understand your problem.
I get only one answer.

We have: . $\frac{\sqrt{x^2-2x} - x}{1}$

Multiply by $\frac{\sqrt{x^2-2x} + x}{\sqrt{x^2-2x} + x}:\;\;$

. . $\frac{\sqrt{x^2-2x} - x}{1} \cdot\frac{\sqrt{x^2-2x} + x}{\sqrt{x^2-2x} + x} \;=\;\frac{(x^2-2x) - x^2}{\sqrt{x^2-2x} + x} \;=\;\frac{-2x}{\sqrt{x^2-2x} + x}$

Divide numerator and denominator by $x\!:$

. . $\dfrac{\dfrac{-2x}{x}}{\dfrac{\sqrt{x^2-2x}}{x} + \dfrac{x}{x}} \;=\; \dfrac{-2}{\dfrac{\sqrt{x^2-2x}}{\sqrt{x^2}} + 1} \;=\;\dfrac{-2}{\sqrt{\dfrac{x^2-2x}{x^2}} + 1}$ Not if x<0, because then $\color{red}x\ne\sqrt{x^2}$ !

. . $=\;\dfrac{-2}{\sqrt{\dfrac{x^2}{x^2}-\dfrac{2x}{x^2}} + 1} \;=\;\dfrac{-2}{\sqrt{1 - \dfrac{2}{x}} + 1}$

Therefore: . $\lim_{x\to\infty}\dfrac{-2}{\sqrt{1 - \frac{2}{x}} + 1} \;=\;\dfrac{-2}{\sqrt{1+0} + 1} \;=\;\frac{-2}{2} \;=\;-1$

The answer x=–1 is correct, of course, for the limit as $x\to\infty$, but not for the limit as $x\to-\infty$.
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