# Thread: Can't find this limit analytically

1. ## 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?

2. ## Re: Can't find this limit analytically

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 $\frac{-1}{4x+4}$

3. ## Re: Can't find this limit analytically

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 ...

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

4. ## 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.)

5. ## Re: Can't find this limit analytically

Originally Posted by Plato
That fraction reduces to $\frac{-1}{4x+4}$
Originally Posted by skeeter
start by simplifying the complex fraction ...

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

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

Yes it is.