# Homework Help: 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

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:
$$\frac{6}{2\sqrt{3}}= \frac{3}{\sqrt{3}}= \sqrt{3}$$