Question

Find the Lim x->0
((1/x+2)-(1/2))/(x)

Answer 1

\[limx-0 \frac{\frac{1}{x+2}-\frac{1}{2}}{x}\]

Answer 2

is that the question?

Answer 3

Yes. Sorry I dont know how to make it look like that yet.

Answer 4

no probs, start by combininb the top fractions, so get common denominators \[\lim x-0 \frac{\frac{2-1(x+2)}{2(x+2)}}{x}\]

Answer 5

okay I have that

Answer 6

Wait..I dont get it. the thing you just posted confused me.

Answer 7

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

Answer 8

yes i had to delete, inversed x accidentally, cancel your -x on top with the bottom \[limx-0 \frac{-1}{2(x+2}\]

Answer 9

\[\LaTeX\]
\[\lim_{x\to a}\]

\lim_{x\to a}

Answer 10

how did you get -1 on the top?

Answer 11

thanks zarkon

Answer 12

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

Answer 13

the x's cancel

Answer 14

okay but where did the (x+2) go?
2-1(x+2)

Answer 15

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

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

Answer 16

\[2-1(x+2)=2-x-2=-x\]

Answer 17

Oooh Okay. I got it now. I accidentaly distributed before. The answer is -1/4 when you plug in the zero right?

Answer 18

yes

Answer 19

This looks somehow like the derived formula of the first principle to me. :P
Ok back to solving.
\[\lim_{x \rightarrow 0} ((1/(x+2))-1/2)/x\]\[=\lim_{x \rightarrow 0} (2-(x+2))/(2(x+2))/x\]\[=\lim_{x \rightarrow 0} (-x)/(2(x+2))/x\]\[=\lim_{x \rightarrow 0} -1/(2x+4)\]\[=-1/4\]

Answer 20

Zarkon: two latex questions if you don't mind:

1. How do you do that canceling out (the diagonal lines on top of the X's) ?
2. How did write that fancy "Latex" in your second post ?

Meanwhile, looking for good reference on Latex (I see many on google).

Thanx!

Answer 21

\[\cancel{Zarkon}\]
\cancel{Zarkon}

\[\LaTeX\]
\LaTeX

Answer 22

Thank you Zarkon - although there was no need to cancel yourself for the demonstration :)