Question

what is the limit of [[1/(3+x)]-(1/3)]/x as x approaches 0?

Answer 1

\[\huge \lim _{x \rightarrow 0} \frac{ \frac 1{3+x} - \frac 13}x\]
that's your question?

Answer 2

yes. it looks a lot prettier now LOL

Answer 3

multiply num and denum by 3(x+3)

Answer 4

combine the fractions in the numerator first..

\[\huge \implies \lim_{x \rightarrow 0} \frac{\frac{3-(3+x)}{3(3+x)}}x\]
do you get how that happened?

Answer 5

yeah

Answer 6

good..so if you simplify that, you get...

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

right?

Answer 7

yes

Answer 8

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

Answer 9

so simplify that further...
\[\huge \implies \lim_{x\rightarrow 0} \quad \frac{-1}{3(3+x)}\]

substitute x to 0 now...what do you get?

Answer 10

maybe - is absent

Answer 11

limit is -1/9

Answer 12

right!!

Answer 13

good job

Answer 14

omg THANKS!! I AM ALMOST DONE WITH TEST CORRECTIONS WOOO

Answer 15

welcome