Calculating $\lim_{x\to+\infty}(\sqrt{x^2-3x}-x)$

Question

Let $f(x) = \sqrt{x^2-3x}$ and $g(x) = x$. Calculate the following limits, showing all working.

I've done the first two -

$$\lim_{x\to0}f(x)\\ =\lim_{x\to0}\sqrt{x^2-3x}=0$$

$$\lim_{x\to-\infty}\frac{f(x)}{g(x)}\\ =\lim_{x\to-\infty}\frac{\sqrt{x^2-3x}}{x}=\sqrt{1-\frac{3}{x}}=1$$

How do I calculate this one?

$$\lim_{x\to+\infty}(f(x)-g(x))$$

could you please verify that the question is not changed by the edit? — lab bhattacharjee, Aug 17 '13 at 09:36
I think the limit the OP is asking about has no multiplication by $,x,$ ... — DonAntonio, Aug 17 '13 at 09:44
For the one I'm trying to work out, it shouldn't be x is approaching NEGATIVE infinity, just infinity. — Katie Malone, Aug 17 '13 at 09:45
Upon checking, the last limit is at $x\to+\infty$, not $x\to-\infty$. (If only users editing posts could be more careful, it seems an awful lot of misprints are introduced this way.) — Did, Aug 17 '13 at 09:46
This post which deals with a more general version of this problem. Many similar limits can be found in the posts which are linked there. — Martin Sleziak, Jan 11 '16 at 13:34

Adriano · Answer 1 · 2013-08-17T09:46:15.873

Your second one is incorrect, since $\sqrt{x^2}=-x$ for all $x<0$. Hence: $$ \lim_{x\to-\infty}\frac{\sqrt{x^2-3x}}{x}=\lim_{x\to-\infty}\frac{\sqrt{x^2-3x}}{-\sqrt{x^2}}=\lim_{x\to-\infty}-\sqrt{1-\frac{3}{x}}=-1 $$

For the last one, we multiply by the conjugate: \begin{align*} \lim_{x \to \infty} [f(x)-g(x)] &= \lim_{x \to \infty} \left[ \sqrt{x^2-3x}-x \right] \\ &= \lim_{x \to \infty} \left[\left(\sqrt{x^2-3x}-x\right) \cdot \dfrac{\sqrt{x^2-3x}+x}{\sqrt{x^2-3x}+x} \right] \\ &= \lim_{x \to \infty} \dfrac{(x^2-3x)-x^2}{\sqrt{x^2-3x}+x} \\ &= \lim_{x \to \infty} \dfrac{-3x}{\sqrt{x^2-3x}+x} \\ &= \lim_{x \to \infty} \dfrac{-3x}{\sqrt{x^2(1-\frac3x)}+x} \\ &= \lim_{x \to \infty} \dfrac{-3x}{x\sqrt{1-\frac3x}+x} \qquad \text{since }\sqrt{x^2}=x \text{ for all }x>0\\ &= \lim_{x \to \infty} \dfrac{-3}{\sqrt{1-\frac3x}+1} \\ &= \dfrac{-3}{\sqrt{1-0}+1} \\ &= \dfrac{-3}{2} \\ \end{align*}

Not unless there is a typo in the question (which seems likely, if you're sure that that's the answer). — Adriano, Aug 17 '13 at 22:48

lab bhattacharjee · Answer 2 · 2013-08-17T09:54:40.923

1

$$\lim_{x\to\infty}\{f(x)-g(x)\}=\lim_{x\to\infty}\{\sqrt{x^2-3x}-x\}$$

$$=\lim_{h\to0}\frac{\sqrt{1-3h}-1}h(\text{ putting } x=\frac1h)$$

Now this can be handled in at least three ways :

Method $1:$

Taylor expansion: $$\lim_{h\to0}\frac{(1-3h)^{\frac12}-1}h=\lim_{h\to0}\frac{1-3h\cdot\frac12+O(h^2)-1}h=-\frac32$$

Method $2:$

L'Hosiptals' Rule :

$$\lim_{h\to0}\frac{(1-3h)^{\frac12}-1}h=\lim_{h\to0}\frac{\frac12\cdot\frac1{\sqrt{1-3h}}\cdot(-3)}1=-\frac32$$

Method $3:$

Rationalizing the numerator like Adriano, $$\lim_{h\to0}\frac{\sqrt{1-3h}-1}h=\lim_{h\to0}\frac{(1-3h)-1}{h(\sqrt{1+3h}+1)}$$

$$=\lim_{h\to0}\frac{-3}{\sqrt{1+3h}+1}(\text{ Cancelling } h\text{ as }h\ne0\text{ as }h\to0 )$$

$$=-\frac32$$

edited Aug 17 '13 at 09:54

answered Aug 17 '13 at 09:31

lab bhattacharjee

274,582

lim [f (x)-g (x)] = [{√(x² -3x)}-x] = x[{(1-(3/x)}^(1/2) -1] = [x(3/2x)+.....higher powers of (1/x)] = 3/2 ? – Katie Malone Aug 17 '13 at 09:33
@KatieMalone, Could you please have a look into the edited answer – lab bhattacharjee Aug 17 '13 at 09:55
There is no negative in the last one I am working on. As I understood, the answer should be 3/2 but I am a little stuck on the working. – Katie Malone Aug 17 '13 at 09:56
@KatieMalone, first of all, I really doubt the existence of $x$ as multiplier in the last question if the answer needs to be finite. Now, if the method is right, so should be the answer. Have you find anything wrong in the three methods? – lab bhattacharjee Aug 17 '13 at 10:02
Yes, you are correct! Would you be able to do me a favor by removing the negatives from your working? And yes! It is not multiplies by x. – Katie Malone Aug 17 '13 at 10:05
@KatieMalone, thanks for your rectification. I think the answer is $-\frac32$. I repeat : "Have you find anything wrong in the three methods?" – lab bhattacharjee Aug 17 '13 at 10:08
I think they look okay to me! – Katie Malone Aug 17 '13 at 10:23

score 0 · Answer 3 · answered Jan 11 '16 at 13:47

To find $\lim \limits_{x \to \infty} \sqrt{x^2-3x}-x:$

Rewrite $\lim \limits_{x \to \infty} \sqrt{x^2-3x}-x=\lim \limits_{x \to \infty} x \left(\sqrt{1-\frac{3}{x}}-1\right)=\lim \limits_{x \to \infty} \frac{\sqrt{1-\frac{3}{x}}-1}{1/x}.$ This gives an indeterminate $\frac{0}{0}$ form. Applying L'Hopital gives us:

$\lim \limits_{x \to \infty} \frac{\left(\frac{3/x^2}{2 \sqrt{1-3/x}}\right)}{(-1/x^2)}=\lim \limits_{x \to \infty} \frac{-3}{2\sqrt{1-\frac{3}{x}}}=\frac{-3}{2}$

score 0 · Answer 4 · answered Jan 11 '16 at 15:25

0

As $$f(x)-g(x)=\sqrt{x^2-3x}-x$$ $$=\frac{x^2-3x-x^2}{\sqrt{x^2-3x}+x}$$ $$=\frac{-3x}{\sqrt{x^2-3x}+x}$$ $$=\frac{-3}{\sqrt{1-3/x}+1}$$ Now when $\lim_{x\to\infty}$ we get $$=\frac{-3}{\sqrt{1-0}+1}$$ $$=\frac{-3}{2}$$

answered Jan 11 '16 at 15:25

Syed Muhammad Asad

485

Calculating $\lim_{x\to+\infty}(\sqrt{x^2-3x}-x)$

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