Limits

**dadon** · February 3rd 2007, 04:30 AM

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 $\frac{(x+1)}{(|x|-1)}$ as $x\rightarrow-1$

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

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

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

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

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

Therefore
$\frac{x+1}{|x|-1}$ with $x=-1+\in$ is $\frac{-1+\in+1}{1-\in-1}=\frac{\in}{-\in}=-1$ 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 $x=-1-\in$ . Hopefully someone can help.

Thank you

Dadon

**CaptainBlack** · February 3rd 2007, 05:47 AM

Originally Posted by dadon

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 $\frac{(x+1)}{(|x|-1)}$ as $x\rightarrow-1$

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

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

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

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

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

Therefore
$\frac{x+1}{|x|-1}$ with $x=-1+\in$ is $\frac{-1+\in+1}{1-\in-1}=\frac{\in}{-\in}=-1$ 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 $x=-1-\in$ . Hopefully someone can help.

Thank you

Dadon

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

RonL

**qpmathelp** · February 3rd 2007, 06:53 AM

Originally Posted by dadon

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 $\frac{(x+1)}{(|x|-1)}$ as $x\rightarrow-1$

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

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

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

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

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

Therefore
$\frac{x+1}{|x|-1}$ with $x=-1+\in$ is $\frac{-1+\in+1}{1-\in-1}=\frac{\in}{-\in}=-1$ 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 $x=-1-\in$ . Hopefully someone can help.

Thank you

Dadon

similar to previous explanation

if h > 0

(a + h) with h-->0 means you approach the point 'a' from the right of 'a' (on the number line)

(a - h) with h-->0 means you approach the point 'a' from the left of 'a' (on the number line)

the corresponding limiting values of the function are the right hand and left hand limits respectively

the function has a limit as x-->a only if the left hand and right hand limits are equal.

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