# Thread: Limits, absolute values, and inequalities with variables in the denominator

1. ## Limits, absolute values, and inequalities with variables in the denominator

Here is the original problem
$\displaystyle$
$\displaystyle \lim_{x\to-1^-}\frac {5} {(x+1)^3}= \infty$

So you do
$\displaystyle \frac {5} {(x+3)}<N$ whenever $\displaystyle -1-\delta <x<-1$

Then you need to solve for x+1 to move on to the next step, but I don't know how...
The book says you should get $\displaystyle \sqrt [3] {5/|N|}$... but I can't figure out how to get the absolute value.

2. Better would be to write $\displaystyle \frac 5{(x+1)^3}<-N$ ($\displaystyle N$ positive) when $\displaystyle -1-\delta<x<-1$ because $\displaystyle \lim_{x\to-1^-}\frac 5{(x+1)^3}=-\infty$.

3. Originally Posted by halbard
Better would be to write $\displaystyle \frac 5{(x+1)^3}<-N$ ($\displaystyle N$ positive) when $\displaystyle -1-\delta<x<-1$ because $\displaystyle \lim_{x\to-1^-}\frac 5{(x+1)^3}=-\infty$.
OK... but that still doesn't answer my question

4. Originally Posted by Chokfull
Here is the original problem
[FONT=monospace]$\displaystyle$
$\displaystyle \lim_{x\to-1^-}\frac {5} {(x+1)^3}= \infty$
The orginal problem is wrong.
$\displaystyle \lim_{x\to-1^-}\frac {5} {(x+1)^3}= {\color{red}-}\infty$

Suppose that $\displaystyle N \in \mathbb{Z}^ +$ then $\displaystyle \sqrt[3]{{\frac{{ - 5}}{N}}} - 1 < x < - 1\, \Rightarrow \,\frac{5}{{\left( {x + 1} \right)^3 }} < - N$.

5. Originally Posted by Plato
The orginal problem is wrong.
$\displaystyle \lim_{x\to-1^-}\frac {5} {(x+1)^3}= {\color{red}-}\infty$

Suppose that $\displaystyle N \in \mathbb{Z}^ +$ then $\displaystyle \sqrt[3]{{\frac{{ - 5}}{N}}} - 1 < x < - 1\, \Rightarrow \,\frac{5}{{\left( {x + 1} \right)^3 }} < - N$.
I think that the problem may have been to prove that this limit is false... that or it was a typo by me or the book... or you are mistaken.
Anyways, does that Z mean positive? if it does, you are wrong--N has to be negative.

6. I'm trying to use the book's method to solve this problem.... The part I need help with is the absolute value in the denominator, I thought there was a rule for inequalities in which you can't multiply each side by the denominator, but I couldn't find it.

7. $\displaystyle \lim_{x\to{a}}f(x)=\infty\Rightarrow$ that for each number $\displaystyle \delta$ such that $\displaystyle |x-a|<\delta$, $\displaystyle \exists$ a number $\displaystyle M$ for which $\displaystyle M<f(x)$.

You can see that the definition is satisfied.However, f(x) goes to NEGATIVE infinity. If you would like to prove it you could choose some arbitraily small number $\displaystyle M$, solve the inequality $\displaystyle M>f(x)$ , then use that value of x (to the left of -1) in the inequality $\displaystyle |x-a|<\delta=\text{slightly larger than the actual value}$ (where a is -1), and show that however close x comes to a, f(x) will still be less than M.

8. Originally Posted by Plato
The orginal problem is wrong.
$\displaystyle x\rightarrow{-1^-},\forall{x}\epsilon(-\infty,-1),\Rightarrow(x+1)^3<0\Rightarrow{f(x)<0}$

You are correct sir.

9. $\displaystyle \frac{5}{(x+1)^{3}}\rightarrow -\infty$ as $\displaystyle x\rightarrow -1^{-}$ if $\displaystyle |x+1|<\delta$ whenever $\displaystyle \frac{5}{(x+1)^{3}}<-N, N>0$ which yields
after solving the absolute value that $\displaystyle \frac{5}{(x+1)^{3}}<\frac{-5}{\delta^{3}}$ so we can write $\displaystyle \frac{-5}{\delta^{3}}=-N$ which directly implies the result.