Help wid limit Question

[umer007]

I need to simplify and solve. Could u plz gimme a step by step answer bcuz i get very confused in these type of cubed root questions.

 

lim ((x+27)^1/3) -3) / x

x->0

 

Its x+27 all under a cubed root subtract 3, all of this divided by x.

 

Thanks in advance

[TD]

Use the fact that [math]\left( {a - b} \right)\left( {a^2 + ab + b^2 } \right) = a^3 - b^3 [/math] and consider the nominator as the factor (a-b) in that expression. Then find out what a and b are and multiply nominator & denominator by the second factor in that expression. The nominator now simplifies to a³-b³ and you'll be able to cancel out an x which will get rid off the indeterminate form. Then it's just filling in

 

I basically said it all, now it's up to you to do the numbers!

[the tree]

[math](x+27)^{\frac{1}{3}}=x^{\frac{1}{3}}+27^{\frac{1}{3}}=x^{\frac{1}{3}}+3[/math]

Then...

[math]\frac{x^{\frac{1}{3}}}{x}[/math]

So...

As [math]x \to \infty, f(x) \to 0[/math]

And as [math]x \to 0, f(x) \to \infty[/math]

[TD]

[math](x+27)^{\frac{1}{3}}=x^{\frac{1}{3}}+27^{\frac{1}{3}}=x^{\frac{1}{3}}+3[/math]

Then...

[math]\frac{x^{\frac{1}{3}}}{x}[/math]

So...

As [math]x \to \infty' date=' f(x) \to 0[/math']

And as [math]x \to 0, f(x) \to \infty[/math]

Eeks!

 

[math]\left( {a + b} \right)^{1/3} \ne a^{1/3} + b^{1/3} [/math] !!!

[Karnage]

[math]

\lim_{x\to\0}

[/math]

 

Now you have to rationalize the numerator and denominator by multiplying the numerator by its conjugate.

[CanadaAotS]

Yah you kinda screwed that up with what TD said...

 

and just to clarify I'll post the limit in latex:

 

[math]\lim_{x\to0} \frac{(x+27)^{\frac{1}{3}} - 3}{x}[/math]

 

I don't understand what TD means, like how to use the [math]a^3 + b^3 [/math] rule with [math]\frac{1}{3}[/math]...

 

since isn't that the difference / sum of cubes rule? 1/3 isn't a perfect cube nor is 3...

[TD]

I'll go into a bit more detail.

 

We would like to get rid off that cube root to simplify the numerator. For square roots, we can use the fact that [math]a^2 - b^2 = \left( {a - b} \right)\left( {a + b} \right)[/math] but for cube roots, that won't help.

 

As I said, we'll be using the identity [math]\left( {a - b} \right)\left( {a^2 + ab + b^2 } \right) = a^3 - b^3[/math]. We consider the current numerator as the factor (a-b) with of course [math]a = \left( {x + 27} \right)^{1/3}[/math] and [math]b = 3[/math]. Now we multiply numerator and denominator with the same factor, namely the second one of our identity, so (a²+ab+b²) with our a and b.

 

After doing that, instead of cancelling these equal factors (then we wouldn't have done a thing...) we can simplify the numerator since the expression there is now equal to a³-b³ according to our identity. But with our a and b, that becomes [math]a^3 - b^3 \Rightarrow \left( {\left( {x + 27} \right)^{1/3} } \right)^3 - 3^3 = x + 27 - 27 = x[/math] and the cube root is gone, just as we wished.

 

The only thing that's left is an x, but that can be cancelled out with the x in the denominator leaving only our added expression in the denominator. It's now possible to simply fill in x = 0 in our limit to find the value.

[CanadaAotS]

I see, I see thanks for that, good to know how to do it

 

btw as dave said the numerator is now [math]x[/math]. The limit now looks like:

 

[math]\lim_{x\to0} \frac{x}{x(a^2 + ab + b^2)} \mbox{ where } a = (x+27)^3 \mbox{ and } b = -3[/math]

 

The x's cancel leaving 1 on the top and we can now substitute a and b:

 

[math]\lim_{x\to0} \frac{1}{(x+27)^{2/3} + 3\cdot(x+27)^{1/3} + 9} [/math]

 

Probably did something wrong as it doesn't look like it will cancel out into something nice... but w/e lol, you can apply the limit now.

 

[math]\frac{1}{27^{2/3} + 3\cdot(27^{1/3}) + 9} [/math]

 

Well it seems 27^(2/3) = 9 so It will be nice The answer is:

 

[math]\frac{1}{27} \mbox{ or if you like it nice and simple... } 3^{-3} [/math]

[TD]

The answer is indeed 1/27 but be careful, b = 3, not -3 in this case (you substituted correctly though).

[CanadaAotS]

oh yah that would've screwed me over... thats the thing with math now, the littlest thing wrong (like say an inverted sign) and the whole thing goes out the window

 

EDIT: by "now" I mean my calculus class lol

[the tree]

Sorry for any confusion caused by my screwed up awsner.

[TD]

Sorry for any confusion caused by my screwed up awsner.

There's no need to apologize, but do you see what your mistake was?

That's quite fundamental, in general (a+b)^n isn't a^n + b^n.

[gibbenergy]

[math]\lim_{x\to0} \frac{(x+27)^{\frac{1}{3}} - 3}{x}[/math]

 

It can be solved quickly if you know derevative. I can post my sols quickly.

Let [math]f(x)=(x+27)^{\frac{1}{3}} - 3[/math] so [math]f(0)=0[/math] then

[math]f'(0)=\lim_{x\to0} \frac{(x+27)^{\frac{1}{3}} - 3}{x-0}[/math]

So you know how to do next?

[Dave]

Well, the idea is to manipulate the limit to get the answer. He's trying to derive the ideas of differentiation from first principles, so doesn't necessarily know anything about the derivative