Question

What is the limit of
(x)÷((1/1+x)−1)
as x approaches 0?

Answer 1

\[\large \lim_{x \rightarrow 0} \frac{ x }{ \left( \frac{ 1 }{ 1+x }-1 \right) }\] this?

Answer 2

yes!

Answer 3

Lets start off by simplifying the denominator you can use the following \[\frac{ a }{ b } \pm \frac{ c }{ d } \implies \frac{ ad \pm bc }{ bd }\]

Answer 4

Go ahead and see what you get

Answer 5

So then would the denominator be \[(1-1+x)/(1+x)\]

Answer 6

wait, the first part should be 1-1-x I think.

Answer 7

\[\huge \lim_{x \rightarrow 0} \frac{ x }{ \left( \frac{ 1-1(1+x) }{ 1+x } \right) } = \lim_{x \rightarrow 0}\frac{ x }{ \left( \frac{ 1-1-x }{ 1+x } \right) }\] correct?

Answer 8

Keep simplifying :)

Answer 9

Yeah. And then I just use the reciprocal of the denominator and plug in the 0, and I get -1?

Answer 10

Bingo!

Answer 11

thank you!!

Answer 12

Yw :)