# Can't find this limit analytically

• Sep 18th 2012, 10:46 AM
Diogenes
Can't find this limit analytically
(((1/(x+1))-(1/4))/(x-3)

I need to find the limit as x approaches 3. I've simplified it to a number of other forms, but I can't get it into a form where x=3 doesn't equate to an undefined value. I know that the answer is -(1/16). Can anyone help point me in the right direction?
• Sep 18th 2012, 11:32 AM
Plato
Re: Can't find this limit analytically
Quote:

Originally Posted by Diogenes
(((1/(x+1))-(1/4))/(x-3)
I need to find the limit as x approaches 3. I've simplified it to a number of other forms, but I can't get it into a form where x=3 doesn't equate to an undefined value. I know that the answer is -(1/16). Can anyone help point me in the right direction?

That fraction reduces to $\frac{-1}{4x+4}$
• Sep 18th 2012, 11:32 AM
skeeter
Re: Can't find this limit analytically
Quote:

Originally Posted by Diogenes
(((1/(x+1))-(1/4))/(x-3)

I need to find the limit as x approaches 3. I've simplified it to a number of other forms, but I can't get it into a form where x=3 doesn't equate to an undefined value. I know that the answer is -(1/16). Can anyone help point me in the right direction?

start by simplifying the complex fraction ...

$\frac{\frac{1}{x+1} - \frac{1}{4}}{x-3} \cdot \frac{4(x+1)}{4(x+1)}$
• Sep 18th 2012, 11:59 AM
SworD
Re: Can't find this limit analytically
Whenever you see a problem like this, try to simplify it to a rational function (quotient of two polynomials, such as $\frac{4 - (x+1)}{4(x-3)(x+1)}$, using various strategies like common denominators, etc. Then pull out any common factors. (You may need to find the roots.)
• Sep 18th 2012, 12:31 PM
Diogenes
Re: Can't find this limit analytically
Quote:

Originally Posted by Plato
That fraction reduces to $\frac{-1}{4x+4}$

Quote:

Originally Posted by skeeter
start by simplifying the complex fraction ...

$\frac{\frac{1}{x+1} - \frac{1}{4}}{x-3} \cdot \frac{4(x+1)}{4(x+1)}$

Quote:

Originally Posted by SworD
Whenever you see a problem like this, try to simplify it to a rational function (quotient of two polynomials, such as $\frac{4 - (x+1)}{4(x-3)(x+1)}$, using various strategies like common denominators, etc. Then pull out any common factors. (You may need to find the roots.)

Thank you all very much. I feel stupid now, but I think I got it. Could you tell me if this reasoning is correct?

$\frac{\frac{1}{x+1} - \frac{1}{4}}{x-3} \cdot \frac{4(x+1)}{4(x+1)}$ = $\frac{4-(x+1)}{4(x-3)(x+1)}$ = $\frac{-(x-3)}{4(x-3)(x+1)}$ = $\frac{-1}{4x+4}$
• Sep 18th 2012, 08:42 PM
SworD
Re: Can't find this limit analytically
Yes it is.