# Help wid limit Question

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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.

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Use the fact that $\left( {a - b} \right)\left( {a^2 + ab + b^2 } \right) = a^3 - b^3$ 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!

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$(x+27)^{\frac{1}{3}}=x^{\frac{1}{3}}+27^{\frac{1}{3}}=x^{\frac{1}{3}}+3$

Then...

$\frac{x^{\frac{1}{3}}}{x}$

So...

As $x \to \infty, f(x) \to 0$

And as $x \to 0, f(x) \to \infty$

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$(x+27)^{\frac{1}{3}}=x^{\frac{1}{3}}+27^{\frac{1}{3}}=x^{\frac{1}{3}}+3$

Then...

$\frac{x^{\frac{1}{3}}}{x}$

So...

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

Eeks!

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

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$\lim_{x\to\0}$

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

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

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

$\lim_{x\to0} \frac{(x+27)^{\frac{1}{3}} - 3}{x}$

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

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

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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 $a^2 - b^2 = \left( {a - b} \right)\left( {a + b} \right)$ but for cube roots, that won't help.

As I said, we'll be using the identity $\left( {a - b} \right)\left( {a^2 + ab + b^2 } \right) = a^3 - b^3$. We consider the current numerator as the factor (a-b) with of course $a = \left( {x + 27} \right)^{1/3}$ and $b = 3$. 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 $a^3 - b^3 \Rightarrow \left( {\left( {x + 27} \right)^{1/3} } \right)^3 - 3^3 = x + 27 - 27 = x$ 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.

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

btw as dave said the numerator is now $x$. The limit now looks like:

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

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

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

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.

$\frac{1}{27^{2/3} + 3\cdot(27^{1/3}) + 9}$

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

$\frac{1}{27} \mbox{ or if you like it nice and simple... } 3^{-3}$

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

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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

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Sorry for any confusion caused by my screwed up awsner.

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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.

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$\lim_{x\to0} \frac{(x+27)^{\frac{1}{3}} - 3}{x}$

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

Let $f(x)=(x+27)^{\frac{1}{3}} - 3$ so $f(0)=0$ then

$f'(0)=\lim_{x\to0} \frac{(x+27)^{\frac{1}{3}} - 3}{x-0}$

So you know how to do next?

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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

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