Limits, absolute values, and inequalities with variables in the denominator

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July 9th 2009, 12:42 PM
Chokfull

Limits, absolute values, and inequalities with variables in the denominator

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

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

Then you need to solve for x+1 to move on to the next step, but I don't know how...(Worried)
The book says you should get $\sqrt [3] {5/|N|}$ ... but I can't figure out how to get the absolute value.
July 9th 2009, 04:58 PM
halbard

Better would be to write $\frac 5{(x+1)^3}<-N$ ( $N$ positive) when $-1-\delta<x<-1$ because $\lim_{x\to-1^-}\frac 5{(x+1)^3}=-\infty$ .
July 13th 2009, 11:42 AM
Chokfull

Quote:

Originally Posted by halbard

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

OK... but that still doesn't answer my question
July 13th 2009, 12:11 PM
Plato

Quote:

Originally Posted by Chokfull

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

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

Suppose that $N \in \mathbb{Z}^ +$ then $\sqrt[3]{{\frac{{ - 5}}{N}}} - 1 < x < - 1\, \Rightarrow \,\frac{5}{{\left( {x + 1} \right)^3 }} < - N$ .
July 13th 2009, 07:22 PM
Chokfull

Quote:

Originally Posted by Plato

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

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

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.(Wondering)
Anyways, does that Z mean positive? if it does, you are wrong--N has to be negative.
July 16th 2009, 01:13 PM
Chokfull

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.
July 16th 2009, 01:32 PM
VonNemo19

http://www.mathhelpforum.com/math-he...8a78af75-1.gif

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

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 $M$ , solve the inequality $M>f(x)$ , then use that value of x (to the left of -1) in the inequality $|x-a|<\delta=\text{slightly larger than the actual value}$ (where a is -1), and show that however close x comes to a, f(x) will still be less than M.
July 16th 2009, 01:41 PM
VonNemo19

Quote:

Originally Posted by Plato

The orginal problem is wrong.

$x\rightarrow{-1^-},\forall{x}\epsilon(-\infty,-1),\Rightarrow(x+1)^3<0\Rightarrow{f(x)<0}$

You are correct sir.
July 16th 2009, 03:45 PM
dr.tea

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