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dwater33 · Sep 22, 2009, 09:43 PM

okay. I did my limits exam and my correctins are due tomorrow and I can't for the life of me figure out how to arrive at the correct answer of this question... the question

limit of x as x approaces 0 of [((1/(x+4))-(1/4))/(x)]

and the answer is -1?16... can somebody explain how to arrive at that, or at least the first step?

Perito · Sep 23, 2009, 05:13 AM

Originally Posted by ;

Limit of x as x approaches 0 of [((1/(x+4))-(1/4))/(x)]

Is this right?

$\lim_{x\to 0}\,\frac {(\frac{1}{x+4} - \frac 14)}{x}$

The first thing to do is to try to simplify the expression so there is only one numerator and one denominator

$\lim_{x\to 0}\, \frac {8-x}{4x^2+16x}$

This doesn't look right because this goes to infinity as x goes to zero. So, let me know what I wrote wrong.

ebaines · Sep 23, 2009, 06:16 AM

Perito - check your math. When you simplify the equation you get this:

$ \frac { \frac 1 {(x+4)} - \frac 1 4 } x = \frac 1 {x(x+4)} - \frac 1 {4x} = \frac {4x-x^2-4x} {4x(x^2+4x)} = \frac {-x^2} {x^2(4x+16)} = \frac {-1} {4x+16} $

This indeed has a limit of -1/16 as x goes to zero.

Perito · Sep 23, 2009, 11:30 AM

Thanks. I was in a hurry.

dwater33 · Sep 23, 2009, 05:44 PM

Originally Posted by ebaines

Perito - check your math. When you simplify the equation you get this:

$ \frac { \frac 1 {(x+4)} - \frac 1 4 } x = \frac 1 {x(x+4)} - \frac 1 {4x} = \frac {4x-x^2-4x} {4x(x^2+4x)} = \frac {-x^2} {x^2(4x+16)} = \frac {-1} {4x+16} $

This indeed has a limit of -1/16 as x goes to zero.

Thank you. For Some reason my brain wasn't working that way. THat makes a lot more sense. :)