# Difficulty to find this limit

1. Apr 22, 2012

1. The problem statement, all variables and given/known data
Hello! I'm having difficulty to find this limit: $$\lim_{x\rightarrow -1}(\sqrt[3]{\frac{x^3+1}{x+1}})$$

2. Relevant equations

3. The attempt at a solution
This is what I'm trying to do to solve this limit: Let $$\frac{x^3+1}{x+1}=u$$ then
$$\lim_{x\rightarrow -1}(\sqrt[3]{u}) = \lim_{x\rightarrow -1}(\sqrt[3]{-1})$$
I know something is wrong I'm just not sure about what is actually wrong. I'm thinking that $$\lim_{x\rightarrow -1}(\sqrt[3]{u})$$ Is wrong because that -1!

Thanks!

Last edited: Apr 22, 2012
2. Apr 22, 2012

### HallsofIvy

Staff Emeritus
All you have done is "hide" the problem in that "u".

I would think that your first step would be to write
$$\frac{x^3+ 1}{x+ 1}= \frac{(x+1)(x^2- x+ 1)}{x+1}= x^2- x+ 1$$

Last edited: Apr 23, 2012
3. Apr 22, 2012

### SammyS

Staff Emeritus
How did you come up with the above result?
Factor x3 + 1 .

4. Apr 22, 2012

Now I can see it!
$$\lim_{x\rightarrow -1}f(x)=3$$ then $$\lim_{x\rightarrow 3}\sqrt[3]{u} =\sqrt[3]{3}$$ and that means $$\lim_{x\rightarrow -1}(\sqrt[3]{\frac{x^3+1}{x+1}})=\sqrt[3]{3}$$
Right?

5. Apr 22, 2012

### e^(i Pi)+1=0

It's asking what the limit is when x approaches -1 not everything inside the radical. Simplify and cancel first then take the limit. Try factoring the denominator.

edit: ^ you are making this way harder than it really is. Forget the limit, use algebra to simplify the expression then plug in -1 like you would in an function and see what you get.

2nd edit: never mind, you got it

6. Apr 22, 2012

### Staff: Mentor

The last expression should be x2 - x + 1.

7. Apr 23, 2012

### HallsofIvy

Staff Emeritus
Thanks! I'm too use to thinking of perfect squares!

8. Apr 23, 2012

### Mentallic

So you have the fraction $$\frac{x^3+1}{x+1}$$ and if you plug x=-1 into this, you get the indeterminate form 0/0. Remember that if you plug a value of x=a into a polynomial and it equates to zero, then it means that you have a root at x=a, and thus you can factor out (x-a) from the polynomial.

This was the first step you needed to realize. Since plugging x=-1 into the numerator and denominator equates both of them to zero, then (x-(-1))=(x+1) is a factor of both the numerator and denominator (but in the denominator it's obvious) so you would then need to figure out how to factor x+1 out of the numerator.

9. Apr 23, 2012

### Whovian

I'd like to say something real quickly. $\displaystyle\lim_{x\to-1}\left(\sqrt[3]{u}\right)\neq\lim_{u\to-1}\left(\sqrt[3]{u}\right)$. By the same arguments, you could show that $\displaystyle\lim_{x\to 0}\left(1\right)$, and, by letting $u=1$, $\displaystyle\lim_{x\to0}\left(u\right)=0$, which is obviously not true.