Question

limit as x tends to 0?

[1/(2+x) - 1/2] / x

Answer 1

use l'hopital's rule. the derivative of the top over the derivative of the bottom. using this we get:

(-1/(2+x)^2)/1

in this limit situation, as x-->0 the limit goes to -1/4

Answer 2

Thanks!

Answer 3

no prob. when the limit looks funky most of the time l'hopital's rule can help out! don't forget it!

Answer 4

\[\frac{\frac{1}{2+x}-\frac{1}{2}}{x}=\frac{\frac{2-(2+x)}{2(2+x)}}{x}=\frac{\frac{-x}{2(x+2)}}{x}=\frac{-x}{2(x+2)}*\frac{1}{x}\]
\[=\frac{-1}{2(x+2)}\]

Answer 5

x->0, f(x)->-1/4