# Rationalizing cubed expressions

1. Nov 19, 2007

### Eezekiel

I know how to rationalize most equations when trying to find limits. However, this problem seem give me trouble.

(((1+x)^3)-1)/x as x approaches 0

I tried the method of multiplying by the conjugate but it doesn't seem to get me anywhere. Mainly its the cubed (1+x) that troubles me.

2. Nov 19, 2007

### varygoode

So you don't want to just expand it traditionally and simplify? You want some sort of easy trick (like multiplying by the conjugate and whatnot)?

3. Nov 19, 2007

### Eezekiel

Could you elaborate and show me how I could expand it traditionally and solve. Maybe I'm just over tired but i can't see how that would work.

4. Nov 19, 2007

### Dick

I don't see anything to rationalize. Just expand (1+x)^3 subtract 1 and divide by x.

5. Nov 19, 2007

### varygoode

Alright, sure. Let's do it in steps.

First, do:

$$(1+x)^2 = (1+x)(1+x)$$

This should be quickly determined to be:

$$1+2x+x^2$$

So, now note:

$$(1+x)^3 = (1+x)(1+x)(1+x) = (1+2x+x^2)(1+x)$$

Then distribute:

$$(1+2x+x^2)(1+x) = (1)(1+x) + (2x)(1+x) + (x^2)(1+x) = 1+x+2x+2x^2+x^2+x^3$$

Then simplify:

$$1+x+2x+2x^2+x^2+x^3 = 1+3x+3x^2+x^3$$

Now, remember your original expression and substitute:

$$\frac{(1+x)^3-1}{x} = \frac{1+3x+3x^2+x^3-1}{x} = \frac{x^3+3x^2+3x}{x} = x^2+3x+3$$

So, then, we have:

$$\lim_{x\rightarrow 0} \left(\frac{(1+x)^3-1}{x}\right) = \lim_{x\rightarrow 0} \left(x^2+3x+3\right)$$

You can take it from here.

Last edited: Nov 19, 2007
6. Nov 20, 2007

### Eezekiel

thank you for clarifying that. Maybe i just needed a quick refresher to show me how again.

7. Nov 21, 2007

### varygoode

Wait, so are you good? Or are you still confused?

8. Nov 21, 2007

### Eezekiel

yes im good thank you