Why Does This Limit Calculation Fail?

[DDarthVader]

Homework Statement

Hello! I'm having difficulty to find this limit: \lim_{x\rightarrow -1}(\sqrt[3]{\frac{x^3+1}{x+1}})

Homework Equations

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!

 

[HallsofIvy]

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

 

[SammyS]

Homework Statement

Hello! I'm having difficulty to find this limit: \lim_{x\rightarrow -1}(\sqrt[3]{\frac{x^3+1}{x+1}})

Homework Equations

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})

How did you come up with the above result?

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!

Factor x³ + 1 .

 

[DDarthVader]

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?

 

[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

 

[Mark44]

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- 2x+ 1

The last expression should be x² - x + 1.

 

[HallsofIvy]

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

 

[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.

 

[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.