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?

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


Is this right?

\Large \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

\Large \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.

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.

Thanks. I was in a hurry.

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. :)