umer007 10 Posted November 9, 2005 Share Posted November 9, 2005 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 Link to post Share on other sites

TD 10 Posted November 9, 2005 Share Posted November 9, 2005 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! Link to post Share on other sites

the tree 222 Posted November 9, 2005 Share Posted November 9, 2005 [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] Link to post Share on other sites

TD 10 Posted November 9, 2005 Share Posted November 9, 2005 [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] !!! Link to post Share on other sites

Karnage 10 Posted November 9, 2005 Share Posted November 9, 2005 [math] \lim_{x\to\0} [/math] Now you have to rationalize the numerator and denominator by multiplying the numerator by its conjugate. Link to post Share on other sites

CanadaAotS 12 Posted November 10, 2005 Share Posted November 10, 2005 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... Link to post Share on other sites

TD 10 Posted November 10, 2005 Share Posted November 10, 2005 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. Link to post Share on other sites

CanadaAotS 12 Posted November 10, 2005 Share Posted November 10, 2005 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] Link to post Share on other sites

TD 10 Posted November 10, 2005 Share Posted November 10, 2005 The answer is indeed 1/27 but be careful, b = 3, not -3 in this case (you substituted correctly though). Link to post Share on other sites

CanadaAotS 12 Posted November 10, 2005 Share Posted November 10, 2005 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 Link to post Share on other sites

the tree 222 Posted November 10, 2005 Share Posted November 10, 2005 Sorry for any confusion caused by my screwed up awsner. Link to post Share on other sites

TD 10 Posted November 11, 2005 Share Posted November 11, 2005 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. Link to post Share on other sites

gibbenergy 10 Posted November 12, 2005 Share Posted November 12, 2005 [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? Link to post Share on other sites

Dave 251 Posted November 12, 2005 Share Posted November 12, 2005 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 Link to post Share on other sites

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