1. ## Confusing Limit problem

Hi, I'm confused on how to figure out the problem $\lim_{x \to 27} \frac{x-27}{x^\frac{1}{3}-3}$. I have the problem worked out, but I am having difficulty understanding how to get from step one to step 2:
$\lim_{x \to 27} \frac {(x^\frac{1}{3})^3-(3)^3}{x^\frac{1}{3}-3}$

to

$\lim_{x \to 27}\frac {(x^\frac{1}{3}-3)(x^\frac{2}{3}+3x^\frac{1}{3}+9)}{(x^\frac{1}{3}-3)}$

Thanks for the help in advance.

2. It was done using the difference of 2 cubes.

$a^3- b^3 = (a - b)(a^2 + ab + b^2)$

3. It is just factorisation.

For example,

If $f(x)=x^{3}-27$

Since $f(3)=0$, by factor theorem, $(x-3)$ is a factor.

So $f(x)=(x-3)(ax^{2}+bx+c)$

Do either long division or comparison to get the other quadratic factor, you will get $f(x)=(x-3)(x^{2}+3x+9)$.

In your limit problem, my $x$ is your $x^{\frac{1}{3}}$

4. $(x^\frac{1}{3})^3-(3)^3$

You can do synthetic division:

$(x^\frac{1}{3})^3+0*x^\frac{2}{3}+0*x^\frac{1}{3}-27$

$x^\frac{1}{3}-3=0$
$x^\frac{1}{3}=3$

So:

3 | 1 0 0 -27
_ |___3 9 27
___ 1 3 9 0

hmm don't know how to do this in latex, nor can i get it to line up . . . sorry.

So you lower the power thing and:

$1*x^\frac{2}{3}+3*x^\frac{1}{3}+9$

hence:
$(x^\frac{1}{3}-3)(x^\frac{2}{3}+3x^\frac{1}{3}+9)$

5. A slightly different way to do this problem would be to "rationalize" the denominator.

Since $a^3- b^3= (a- b)(a^2+ ab+ b^2)$, [tex](x^{1/3}- 3)(x^{2/3}+ 3x^{1/3}+ 9)= x- 27. Multiplying both numerator and denominator by $x^{2/3}+ 3x^{1/3}+ 9$ gives
$\frac{(x- 27)(x^{2/3}+ 3x^{1/3}+ 9)}{x- 27}= x^{2/3}+ 3x^{1/3}+ 9$. Now you can let x= 27 in that.

6. Originally Posted by cynlix

Hi, I'm confused on how to figure out the problem $\lim_{x \to 27} \frac{x-27}{x^\frac{1}{3}-3}$.
put $t=x^{\frac13}$ and the limit becomes $\underset{t\to 3}{\mathop{\lim }}\,\frac{t^{3}-27}{t-3},$ which you know how to solve.