# Thread: Evaluating Limits with rationals.

1. ## Evaluating Limits with rationals.

I got a question on my test today and I spent a great deal of time on the question.
lim as x -> 27 (27-x)/[(x^1/3)-3]

I was able to approximate the limit by evaluating left and right of 27; however, there was another method where a change of variable was used, for example let u=(x^1/3).

I was unsure how to proceed this way. any help would be appreciated.

2. Originally Posted by darksoulzero
I got a question on my test today and I spent a great deal of time on the question.
lim as x -> 27 (27-x)/[(x^1/3)-3]

I was able to approximate the limit by evaluating left and right of 27; however, there was another method where a change of variable was used, for example let u=(x^1/3).

I was unsure how to proceed this way. any help would be appreciated.
Let $\displaystyle \displaystyle {u = x^{1/3} \Rightarrow x = u^3}$. Then your limit becomes $\displaystyle \displaystyle{\lim_{u \to 3} \frac{27 - u^3}{u - 3}}$ which should be simple for you to do.

3. More help:

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

4. Originally Posted by darksoulzero
I got a question on my test today and I spent a great deal of time on the question.
lim as x -> 27 (27-x)/[(x^1/3)-3]

I was able to approximate the limit by evaluating left and right of 27; however, there was another method where a change of variable was used, for example let u=(x^1/3).

I was unsure how to proceed this way. any help would be appreciated.
You can factor out $\displaystyle 3-x^{1/3}$ in the numerator and then cancel that factor against the denominator like this:

$\displaystyle \lim\limits_{x\to 27}\frac{27-x}{x^{1/3}-3}=\lim\limits_{x\to 27}\frac{(3-x^{1/3})(3^2+3x^{1/3}+x^{2/3})}{x^{1/3}-3}=\lim\limits_{x\to 27}\left[-(3^2+3x^{1/3}+x^{2/3})\right]=-27$

5. Originally Posted by Failure
You can factor out $\displaystyle 3-x^{1/3}$ in the numerator and then cancel that factor against the denominator like this:

$\displaystyle \lim\limits_{x\to 27}\frac{27(27-x)}{x^{1/3}-3}=\lim\limits_{x\to 27}\frac{27(3-x^{1/3})(3^2+3x^{1/3}+x^{2/3})}{x^{1/3}-3}=\lim\limits_{x\to 27}\left[-27(3^2+3x^{1/3}+x^{2/3})\right]=-729$
I think you'll find that there's a mistake somewhere here because -729 is not the answer.

6. Originally Posted by mr fantastic
I think you'll find that there's a mistake somewhere here because -729 is not the answer.
yeah, the answer is actually -27, but thanks for assistance gentlemen I did not realize that you could have factored out the -27.

7. Originally Posted by mr fantastic
I think you'll find that there's a mistake somewhere here because -729 is not the answer.
Thank you for the hint: I had mistakenly added the factor 27 . I will fix that mistake in my reply to fit the original problem statement (although, what I had written ist not, strictly speaking, wrong: it is just not the answer to the original problem).

8. Yeha you're right it wasn't wrong.