1. ## Evaluating limits.

find the given limit as x approaches 0.
([1/(x+1)] /x) (-1/x)

thanks,
zirr

2. Originally Posted by Zirrick
find the given limit as x approaches 0.
([1/(x+1)] /x) (-1/x)

thanks,
zirr
Is this what you are tying to solve:

$\displaystyle\lim_{x\to 0}\left[\frac{\frac{1}{x+1}}{x}\cdot\frac{-1}{x}\right]$

3. Originally Posted by dwsmith
Is this what you are tying to solve:
$\displaystyle\lim_{x\to 0}\left[\frac{\frac{1}{x+1}}{x}\cdot\frac{-1}{x}\right]$
If $\displaystyle\left[\frac{\frac{1}{x+1}}{x}\cdot\frac{-1}{x}\right]$ is correct then note that is equal to
$\displaystyle\left[{\frac{x}{x+1}\cdot\frac{-1}{x}\right]=\frac{-1}{x+1}$

4. Originally Posted by dwsmith
Is this what you are tying to solve:

$\displaystyle\lim_{x\to 0}\left[\frac{\frac{1}{x+1}}{x}\cdot\frac{-1}{x}\right]$
um, not exactly i think i wrote it wrong. maybe like this?
[(1/x+1)-1/x]

5. Originally Posted by Zirrick
um, not exactly i think i wrote it wrong. maybe like this?
[(1/x+1)-1/x]
$\displaystyle\frac{1}{x+1}-\frac{1}{x}=\frac{x-x-1}{x(x+1)}=\frac{-1}{x(x+1)}$

$\displaystyle\lim_{x\to 0}\frac{-1}{x(x+1)}$

Now, is this correct?

6. ## I'm sorry I'm not to familiar with how exactly to write this.

[(1/(x+1))-1]/x

thanks for trying to help.

7. Originally Posted by Zirrick
[(1/(x+1))-1]/x

thanks for trying to help.
$\displaystyle\frac{\frac{1}{x+1}-1}{x}\text{?}$

This one makes sense since it utilizes L'Hopitals Rule.

8. Yes! thats the exact way its written!
Thanks, what are the possible ways to find the limit? as x approaches 0.

9. Originally Posted by Zirrick
Yes! thats the exact way its written!
Thanks, what are the possible ways to find the limit? as x approaches 0.
$\displaystyle\lim_{x\to 0}\left[\frac{\frac{1}{x+1}-1}{x}\right]=\frac{0}{0}$

$\displaystyle\lim_{x\to 0}\frac{f'(x)}{g'(x)}\Rightarrow\lim_{x\to 0}\frac{\frac{-1}{(x+1)^2}}{1}=-1$

10. can you explain how to use L'Hopital's rule?

11. Originally Posted by Zirrick
is there other way to do this?
or is this the only way?
$\displaystyle\frac{\frac{1}{x+1}-1}{x}=\frac{\frac{1-x-1}{x+1}}{x}=\frac{\frac{-x}{x+1}}{x}=\cdots$

After simplification, take the limit.

12. Originally Posted by Zirrick
can you explain how to use L'Hopital's rule?
When you have an indeterminant form, you can take the derivative of both the numerator, denominator, and then check the limit.

13. Originally Posted by Plato
If $\displaystyle\left[\frac{\frac{1}{x+1}}{x}\cdot\frac{-1}{x}\right]$ is correct then note that is equal to
$\displaystyle\left[{\frac{x}{x+1}\cdot\frac{-1}{x}\right]=\frac{-1}{x+1}$
'
No, it isn't. $\frac{\frac{1}{x+1}}{x}= \frac{1}{x+1}\frac{1}{x}= \frac{1}{x(x+1)}$
NOT $\frac{x}{x+1}$. That would $\frac{1}{\frac{x+1}{x}}$.

A very strange error for Plato to make. (What were you smoking? I want some!)

14. Originally Posted by dwsmith
When you have an indeterminant form, you can take the derivative of both the numerator, denominator, and then check the limit.
If f(x)-> 0 and g(x)-> 0 as x->a, then [tex]\lim_{x\to a}\frac{f(x)}{g(x)}= \lim_{x\to a}\frac{f'(x)}{g'(x)}[/itex]
and you can extend to $\frac{\infty}{\infty}$ and other "indeterminant forms".

Note that numerator and denominator are differentiated separately, you are not differntiating the fraction using the quotient law.