# Homework Help: Simple limit (verification)

1. Sep 19, 2010

### seto6

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

2. Relevant equations

N/A (L'Hospital (maybe)??

3. The attempt at a solution

so this is what i tried,

since i got infinity - infininty, i tried to convert to a fraction then went on from there to get 2

but using wolfram it gave me -2, mind that it used l'hoapital after converting to a fraction, i didn't find it necessary to apply L'H, since i simplified and didn't get an intermidate form.

wolfram:: http://www.wolframalpha.com/input/?i=lim%20x-%3E%20-infinity%20%28sqrt%28%28x^2%29%2B2x%29-%28sqrt%28%28x^2%29-2x%29&t=wvg01

#### Attached Files:

• ###### 1z6uzp1.png
File size:
707 bytes
Views:
56
2. Sep 19, 2010

### ╔(σ_σ)╝

Your last step is not really justified.

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

Since x gets more and more negative

|x| = -x ( in this case)

3. Sep 19, 2010

### seto6

abs(x)abs so i can use it as -x and x and see if they boht approch the same #?

4. Sep 19, 2010

### ╔(σ_σ)╝

In your specific case they will not approach the same number as you can see. That is why wolfram got -2.

|x| = -x if x <0
|x| = x if x >= 0

In you case x was going to negative infinity. So it only make sense to put abs(x) = -x.
In your solution you simiply put $$|x| = x$$ which is not justified since you are bringing stuff out of the square root.

5. Sep 21, 2010

### seto6

hey.. so how can i work with the absolute's? help...

6. Sep 21, 2010

### ╔(σ_σ)╝

What do you mean ?

7. Sep 21, 2010

### seto6

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

from here can i find the limit like i take in to account like split into -x and x and find the values... i did find them but they are -2 and 2... where im i going wrong

EDIT: you said something about using -x...why is that...

8. Sep 22, 2010

### HallsofIvy

I don't see how that will do any good at all since you still have that "|x|" going to infinity and the difference going to 0.

What I recommend, even though you don't really have a fraction, is to "rationalize the numerator":
$$\sqrt{x^2+ 2x}- \sqrt{x^2- 2x}= \left(\sqrt{x^2+ 2x}- \sqrt{x^2- 2x}\right)\frac{\sqrt{x^2+ 2x}+ \sqrt{x^2- 2x}}{\sqrt{x^2+ 2x}+ \sqrt{x^2- 2x}}$$
$$= \frac{x^2+ 2x- (x^2- 2x)}{\sqrt{x^2+ 2x}+ \sqrt{x^2- 2x}}$$
$$= \frac{4x}{\sqrt{x^2+ 2x}+ \sqrt{x^2- 2x}}$$

Now divide both numerator and denominator by x, making it $x^2$ inside the square roots:
$$\frac{4}{\sqrt{1+ 2/x}+ \sqrt{1- 2/x}}$$

Now take the limit as x goes to negative infinity.

9. Sep 22, 2010

### ╔(σ_σ)╝

Okay I will explain in more details.

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

This is all good. You did that part excellently.

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

Take the $$x^2$$ term out
$$=\frac{4x}{ \left|x \right| \left( \sqrt{1+ 2/x}+ \sqrt{1- 2/x} \right)}$$

The absolute value sign is because...
$$\sqrt{x^2} = \left|x \right|$$

We know that |x| is defined as follows

|x| = -x if x<0
|x| = x if x >= 0

In your limit x <0 since you are making x approach negative infinity . x would be negative so -x would make x positive. Do you understand this ?

And that is why you should replace |x| with -x

That is ...
$$=\frac{4x}{-x \left( \sqrt{1+ 2/x}+ \sqrt{1- 2/x} \right)}$$
$$=\frac{-4}{ \sqrt{1+ 2/x}+ \sqrt{1- 2/x}}$$

And now you can happily take your limit to and you will see that it approaches $$\frac{-4}{2} = -2$$

That is exactly what OP did which was incorrect.

Following your solution leads to a limit going to $$\frac{4}{2} =2$$ which is not correct!

10. Sep 22, 2010

### seto6

╔(σ_σ)╝ thank you so much i see where i went wrong...(I always make these mistake)