Limits question

1. Jul 29, 2008

JFonseka

1. The problem statement, all variables and given/known data

L = lim x-> infinity ($$\sqrt{3x^{2}+4x}$$ - $$\sqrt{3x^{2}-2x}$$

2. Relevant equations
None

3. The attempt at a solution

I first multiplied the whole equation by $$\sqrt{1}$$ to get rid of the square root.
Thus I got (3x^2 + 4x - (3x^2 - 2x)). With this, I divided the terms by x to get ultimately 4 + 2 = 6, but I don't think this answer is right because I didn't really substitute x anywhere. It just turned out to be some random algebra at the end.

Thoughts?

2. Jul 29, 2008

moemoney

You cant just square everything to get rid of the square roots.
(a + b)^2 does not equal to a^2 + b^2.

3. Jul 29, 2008

Dick

To simplify something like sqrt(A)-sqrt(B) multiply by (sqrt(A)+sqrt(B))/(sqrt(A)+sqrt(B)) i.e. '1' to get (A-B)/(sqrt(A)+sqrt(B)). You seem to be losing the denominator.

4. Jul 29, 2008

JFonseka

But doing that doesn't really help me because now I have square roots on the denominator.

5. Jul 29, 2008

JFonseka

Trying some factoring out, I got

6x/(x sqrt(3 + 1/4x) + x sqrt(3 + 1/2x))

and got stuck

6. Jul 29, 2008

JFonseka

Ah ha, I think I got it now.

Using l'hopital's rule I think? dividing by x, and therefore getting 6/2sqrt(3) as x heads to infinity in 1/4x and 1/2x and thus becoming 0 there.

Thanks to Dick and others who helped.

6/2sqrt(3)

7. Jul 29, 2008

moemoney

Yup thats the correct answer. Or in simpler terms its sqrt(3). I dont think l'hopitals rule should be used here.

8. Jul 29, 2008

HallsofIvy

Staff Emeritus
It's certainly not neccesary to use L'Hopital's rule.

JFonseka, do exactly what you were told to do initially: multiply $\sqrt{3x^2+ 4x}- \sqrt{3x^2- 2x}$ by
$$\frac{\sqrt{3x^2+ 4x}+ \sqrt{3x^2- 2x}}{\sqrt{3x^2+ 4x}+ \sqrt{3x^2- 2x}}$$
to get
$$\frac{3x^2+ 4x- 3x^2+ 2x}{\sqrt{3x^2+ 4x}+ \sqrt{3x^2- 2x}}$$
$$\frac{6x}{\sqrt{3x^2+ 4x}+ \sqrt{3x^2- 2x}}$$

Yes, you now have square roots in the denominator but they are added so the "$\infty- \infty$" problem that you had initially doesn't occur here. Since
Do the "standard" method when dealing with limits at infinity: divide both numerator and denominator by the highest power of x- here just x itself. Of course, the "x" you divide by in the denominator become "x2" in the square root.
Doing that you have:
$$\frac{6}{\sqrt{3+ 4/x}+ \sqrt{3- 2/x}}$$

Now, as x goes to infinity, those fractions with x in the denominator go to 0. The limit is:
$$\frac{6}{2\sqrt{3}}= \frac{3}{\sqrt{3}}= \sqrt{3}$$