## anonymous one year ago What is the limit of (x)÷((1/1+x)−1) as x approaches 0?

1. anonymous

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

2. anonymous

yes!

3. anonymous

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

4. anonymous

Go ahead and see what you get

5. anonymous

So then would the denominator be $(1-1+x)/(1+x)$

6. anonymous

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

7. anonymous

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

8. anonymous

Keep simplifying :)

9. anonymous

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

10. anonymous

Bingo!

11. anonymous

thank you!!

12. anonymous

Yw :)