Question

Find the limit (if it exists).
lim x->0 [[1/(3+x)]-(1/3)]/x

Answer 1

\(\huge\frac{1}{3+x}-\frac{1}{3}=\frac{3-(3+x)}{3(3+x)}=??\)

Answer 2

can u solve further?

Answer 3

yes, thank you!

Answer 4

welcome :)
let me know your final answer,so that we can verify.

Answer 5

I get 0 but my book says -1/9?

Answer 6

u have:
\(\huge\frac{1}{3+x}-\frac{1}{3}=\frac{3-(3+x)}{3(3+x)}=\frac{-x}{3x(3+x)}\)
right ?

Answer 7

No, I thought I could just solve by substituting 0 for x.

Answer 8

I dont think you notcied that the orignal equation was divided by x.

Answer 9

but u have x in denominator also , right ?
so u just cannot put x=0
cancel out x from numerator and denominator
then put x=0

Answer 10

Oh, i see now. thank you.

Answer 11

did u get -1/9 ?

Answer 12

Yes. But could you please explain how you got from 3-(3+x)/3(3+x) to -x/3x(3+x)? I was a bit confused about that.

Answer 13

the numerator, right ?
3-(3+x) = 3-3-x = 0-x = -x
ok?
denominator is same.

Answer 14

thank you

Answer 15

welcome :)