# Limits

• Feb 3rd 2007, 04:30 AM
Limits
Hi everyone!

I started revision today and im stuck on a limit question. I have the solution but I still have no clue of how to solve it. maybe another simple example may help.

Question 1.
Evaluate the limit of $\displaystyle \frac{(x+1)}{(|x|-1)}$ as $\displaystyle x\rightarrow-1$

If $\displaystyle x=-1$ then get $\displaystyle \frac{-1+1}{1-1}=\frac{0}{0}!$

Therefore let $\displaystyle x=-1-\in$ then $\displaystyle \in\rightarrow0$ in limit $\displaystyle (\in >0)$

i.e. $\displaystyle \frac{-1-\in+1}{(1+\in)-1} = \frac{-\in}{\in}=-1$

$\displaystyle x=-1-\in=-(1+\in)$
$\displaystyle |-1-\in|=(1+\in)$

If let $\displaystyle x=-1+\in=-(1-\in)$ and $\displaystyle |-1+\in|=1-\in$ $\displaystyle (\in>0)$

Therefore
$\displaystyle \frac{x+1}{|x|-1}$ with $\displaystyle x=-1+\in$ is $\displaystyle \frac{-1+\in+1}{1-\in-1}=\frac{\in}{-\in}=-1$ same answer as before

Thats the answer above. I think they use two solutions but both ways are right. But I don't get why you have to put in $\displaystyle x=-1-\in$. Hopefully someone can help.

Thank you

• Feb 3rd 2007, 05:47 AM
CaptainBlack
Quote:

Hi everyone!

I started revision today and im stuck on a limit question. I have the solution but I still have no clue of how to solve it. maybe another simple example may help.

Question 1.
Evaluate the limit of $\displaystyle \frac{(x+1)}{(|x|-1)}$ as $\displaystyle x\rightarrow-1$

If $\displaystyle x=-1$ then get $\displaystyle \frac{-1+1}{1-1}=\frac{0}{0}!$

Therefore let $\displaystyle x=-1-\in$ then $\displaystyle \in\rightarrow0$ in limit $\displaystyle (\in >0)$

i.e. $\displaystyle \frac{-1-\in+1}{(1+\in)-1} = \frac{-\in}{\in}=-1$

$\displaystyle x=-1-\in=-(1+\in)$
$\displaystyle |-1-\in|=(1+\in)$

If let $\displaystyle x=-1+\in=-(1-\in)$ and $\displaystyle |-1+\in|=1-\in$ $\displaystyle (\in>0)$

Therefore
$\displaystyle \frac{x+1}{|x|-1}$ with $\displaystyle x=-1+\in$ is $\displaystyle \frac{-1+\in+1}{1-\in-1}=\frac{\in}{-\in}=-1$ same answer as before

Thats the answer above. I think they use two solutions but both ways are right. But I don't get why you have to put in $\displaystyle x=-1-\in$. Hopefully someone can help.

Thank you

You use $\displaystyle x=1\pm \epsilon, \epsilon > 0$ as you want to look at the behaviour of the expression near $\displaystyle x=-1$, and you need to handle the $\displaystyle \pm \epsilon$ cases seperatly to control the behaviour of $\displaystyle |1 \pm \epsilon|$ while looking at the limit from above and below.

RonL
• Feb 3rd 2007, 06:53 AM
qpmathelp
Quote:

Hi everyone!

I started revision today and im stuck on a limit question. I have the solution but I still have no clue of how to solve it. maybe another simple example may help.

Question 1.
Evaluate the limit of $\displaystyle \frac{(x+1)}{(|x|-1)}$ as $\displaystyle x\rightarrow-1$

If $\displaystyle x=-1$ then get $\displaystyle \frac{-1+1}{1-1}=\frac{0}{0}!$

Therefore let $\displaystyle x=-1-\in$ then $\displaystyle \in\rightarrow0$ in limit $\displaystyle (\in >0)$

i.e. $\displaystyle \frac{-1-\in+1}{(1+\in)-1} = \frac{-\in}{\in}=-1$

$\displaystyle x=-1-\in=-(1+\in)$
$\displaystyle |-1-\in|=(1+\in)$

If let $\displaystyle x=-1+\in=-(1-\in)$ and $\displaystyle |-1+\in|=1-\in$ $\displaystyle (\in>0)$

Therefore
$\displaystyle \frac{x+1}{|x|-1}$ with $\displaystyle x=-1+\in$ is $\displaystyle \frac{-1+\in+1}{1-\in-1}=\frac{\in}{-\in}=-1$ same answer as before

Thats the answer above. I think they use two solutions but both ways are right. But I don't get why you have to put in $\displaystyle x=-1-\in$. Hopefully someone can help.

Thank you