Rationalizing cubed expressions

[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.

 

[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)?

 

[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.

 

[Dick]

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

 

[varygoode]

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.

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

First, do:

[tex] (1+x)^2 = (1+x)(1+x)[/tex]

This should be quickly determined to be:

[tex] 1+2x+x^2 [/tex]

So, now note:

[tex] (1+x)^3 = (1+x)(1+x)(1+x) = (1+2x+x^2)(1+x) [/tex]

Then distribute:

[tex] (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 [/tex]

Then simplify:

[tex] 1+x+2x+2x^2+x^2+x^3 = 1+3x+3x^2+x^3 [/tex]

Now, remember your original expression and substitute:

[tex] \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 [/tex]

So, then, we have:

[tex] \lim_{x\rightarrow 0} \left(\frac{(1+x)^3-1}{x}\right) = \lim_{x\rightarrow 0} \left(x^2+3x+3\right) [/tex]

You can take it from here.

 

[Eezekiel]

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

 

[varygoode]

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

 

[Eezekiel]

yes I am good thank you