Limits

**darkblue** · September 20th 2009, 05:31 PM

Can someone please help me? walk me through this?

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

**VonNemo19** · September 20th 2009, 05:35 PM

Originally Posted by darkblue

Can someone please help me? walk me through this?

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

First of all, we should verify our associations.

Is this what you meant?

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

**darkblue** · September 20th 2009, 05:42 PM

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

**VonNemo19** · September 20th 2009, 05:54 PM

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?

**darkblue** · September 20th 2009, 05:58 PM

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....

**superdude** · September 20th 2009, 06:02 PM

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

**Chris L T521** · September 20th 2009, 06:05 PM

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

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

L'Hôpital's rule is overkill for this one.

**skeeter** · September 20th 2009, 06:05 PM

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?

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

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

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

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

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

**darkblue** · September 20th 2009, 06:19 PM

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?

**VonNemo19** · September 20th 2009, 06:22 PM

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.

**darkblue** · September 20th 2009, 06:30 PM

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?

**VonNemo19** · September 20th 2009, 06:31 PM

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?

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