1. ## Limits

lim (1/x) -1/(x^2)+x)
x->0

2. Originally Posted by darkblue

lim (1/x) -1/(x^2)+x)
x->0
First of all, we should verify our associations.

Is this what you meant?

$\lim_{x\to0}\left(\frac{1}{x}-\frac{1}{x^2+x}\right)$

3. Originally Posted by VonNemo19
First of all, we should verify our associations.

Is this what you meant?

$\lim_{x\to0}\left(\frac{1}{x}-\frac{1}{x^2+x}\right)$
yes that exactly

4. Originally Posted by darkblue
yes that exactly
Begin by combining the fractions by finding the LCD and note the indeterminate form $\frac{0}{0}$.

Do you know how to proceed?

5. Originally Posted by VonNemo19
Begin by combining the fractions by finding the LCD and note the indeterminate form $\frac{0}{0}$.

Do you know how to proceed?
ok so i got (x^2)-2x all over (x^3)-(x^2)

now i really dont know what to do....

6. use l'hospital's rule. I advise you to look it up in your textbook or google it.

7. Originally Posted by darkblue
ok so i got (x^2)-2x all over (x^3)-(x^2)

now i really dont know what to do....
That's not quite right.

First note that $\frac{1}{x}-\frac{1}{x^2+x}=\frac{1}{x}-\frac{1}{x(x+1)}$.

The LCD is $x(x+1)$. Can you try to combine the fractions now? You should see some kind of nice cancellation occur.

Originally Posted by superdude
L'Hôpital's rule is overkill for this one.

8. Originally Posted by darkblue
ok so i got (x^2)-2x all over (x^3)-(x^2)

now i really dont know what to do....
forget how to combine fractions?

$\frac{1}{x} - \frac{1}{x^2+x}$

$\frac{1}{x} - \frac{1}{x(x+1)}$

$\frac{x+1}{x(x+1)} - \frac{1}{x(x+1)}$

$\frac{(x+1) - 1}{x(x+1)}$

$\frac{x}{x(x+1)}$

9. Originally Posted by skeeter
forget how to combine fractions?

$\frac{1}{x} - \frac{1}{x^2+x}$

$\frac{1}{x} - \frac{1}{x(x+1)}$

$\frac{x+1}{x(x+1)} - \frac{1}{x(x+1)}$

$\frac{(x+1) - 1}{x(x+1)}$

$\frac{x}{x(x+1)}$

okay thankyou. but now after doing that and we are left with
$\frac{x}{x(x+1)}$ what do i do now?

10. Originally Posted by darkblue
okay thankyou. but now after doing that and we are left with
$\frac{x}{x(x+1)}$ what do i do now?
Cancel the x's and substitute directly.

11. Originally Posted by VonNemo19
Cancel the x's and substitute directly.
ok thank you..so i get 1/(x+1) substitute zero and get 1/1 = 1

so is the limit 1 then?

12. Originally Posted by darkblue
ok thank you..so i get 1/(x+1) substitute zero and get 1/1 = 1

so is the limit 1 then?