[Q] Limiting Condition SUm

[theblackalchemist]

[math]

\text{Prove That..}

Lim_{(x->0)}{(1+x)^{(1/3)}/x-(1-x)^{(1/3)}}/x = 2/3

[/math]

 

I've been tryiing to figure this out.

maybe i'm missing something trivial? :shrug:

Thanks

[theblackalchemist]

i'm also trying to figure out how to get the "divided by" sign to the whole fraction :shrug:

[CraigD]

You can solve problems like this using l'Hôpital's rule,

[math]\lim_{x \to c} \frac{f(x)}{g(x)} = \lim_{x \to c} \frac{f'(x)}{g'(x)}[/math]

 

as follows:

Given:

[math]f(x)= (1+x)^{\frac13} - (1-x)^{\frac13}[/math]

 

[math]g(x)= x[/math]

 

Taking the first derivative with the power rule:

[math]f'(x)= \frac13(1+x)^{-\frac23} +\frac13(1-x)^{-\frac23}[/math]

 

[math]g'(x)= 1[/math]

 

Then use l'Hôpital's rule:

[math]\lim_{x \to 0} \frac{(1+x)^{\frac13} - (1-x)^{\frac13}}{x} = \lim_{x \to 0} \frac{\frac13(1+x)^{-\frac23} +\frac13(1-x)^{-\frac23}}{1} = \frac{\frac13 +\frac13}{1} = \frac23[/math]

i'm also trying to figure out how to get the "divided by" sign to the whole fraction

You can render a fraction using the \frac{}{} LaTeX:math element. Click the reply tag on this post to see an example.

[theblackalchemist]

I tried it like this

[math]

 

 

\frac{\sqrt[3]{\frac{1+x}{x^3}}-\sqrt[3]{\frac{1+x}{x^3}}}{x/x}

 

 

 

[/math]

 

then factorizing the denom, we'll get 1.

 

however i'm stuck on factorizing the numerator.

 

 

the hopital's rule is fine :)

Thanks

TBA