Question

Complex fractions...

Answer 1

\[\left[\frac{ 1 }{ 3+x }-\frac{ 1 }{ 3 }\right] \over x\]

Answer 2

add the top, what do you get?

Answer 3

don't you have to have like denominators to add the top??? and i cant figure out how to make them the same

Answer 4

well, preferably, yes, but when you don't have such, you can always make the LCD just the product of all the denominators

Answer 5

won't that make the numerator zero? if it helps, i'm trying to evaluate the liit of this as it approaches zero.

Answer 6

limit*

Answer 7

\(\bf \cfrac{
\frac{ 1 }{ 3+x }-\frac{ 1 }{ 3 }
}{
x
} \implies
\cfrac{
\frac{ 3-(3+x) }{3(3+x) }
}{x}\implies
\cfrac{
\frac{ 3-3-x }{3(3+x) }
}{x}\implies \cfrac{ 3-3-x }{3(3+x) } \times \cfrac{1}{x}\)


recalll that \(\bf \cfrac{\frac{a}{b}}{\frac{c}{d}} \implies \cfrac{a}{b}\times \cfrac{d}{c}\)

Answer 8

well, you never said anything about limit :|

Answer 9

but anyhow , the same applies