Question

I need to find the limit of the Function ((1/3+x)-(1/3))/x as x aproches 0

Answer 1

what are you allowed to use?

Answer 2

\[\frac{\frac{1}{3}+x-\frac{1}{3}}{x}\] like this?

Answer 3

\[\frac{\frac{1}{3+x}-\frac{1}{3}}{x}?\]

Answer 4

ooh, that makes more sense

Answer 5

lol

Answer 6

go ahead. and when you are done, tell me where i can find those nice fan testimonials

Answer 7

yes Myininaya thats it

Answer 8

\[\lim_{x \rightarrow 0}\frac{\frac{1}{3+x}-\frac{1}{3}}{x} \cdot \frac{(3+x)(3)}{(3+x)(3)}\]

Answer 9

\[\lim_{x \rightarrow 0}\frac{3-(3+x)}{3x(3+x)}\]

Answer 10

\[\lim_{x \rightarrow 0}\frac{-x}{3x(3+x)}\]

Answer 11

\[\lim_{x \rightarrow 0}\frac{- \not{x}}{3 \not {x} (3+x)}=\lim_{x \rightarrow 0}\frac{-1}{3(3+x)}\]

Answer 12

now plug in zero

Answer 13

\[\frac{-1}{3(3+0)}=\frac{-1}{9}\]

Answer 14

Thank you so much and sorry about the equation confusion I just started using the site

Answer 15

satellite you want to see your fan testimonials or do you want to leave others a fan testimonial?
just go to your profile or someone's to read theirs or to leave one

Answer 16

its cool vinder

Answer 17

did you follow everything i did vinder?

Answer 18

oh so now you want to clear the fractions. before you did the arithmetic. i will do it that way
\[\frac{\frac{3-(3+x)}{(3+x)3}}{x}\]
\[\frac{-x}{(3+x)3x}\]
\[\frac{-1}{(3+x)3}\] and
\[\lim_{x\rightarrow }\frac{-1}{(x+3)3}=\frac{-1}{3\times 3}=-\frac{1}{9}\]

Answer 19

satellite combine the top fractions first which is fine too

Answer 20

that is what you used to do i remember. i want to see them and leave them both!

Answer 21

Yeah I forgot how to simplify multiple fractions in the numerator thank you so much

Answer 22

yes i used to do it that way but i like this way better

Answer 23

i like it better for clearing complex fractions but for some reason here i like the artithmetic