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

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

Mathematics
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\[\large \lim_{x \rightarrow 0} \frac{ x }{ \left( \frac{ 1 }{ 1+x }-1 \right) }\] this?
yes!
Lets start off by simplifying the denominator you can use the following \[\frac{ a }{ b } \pm \frac{ c }{ d } \implies \frac{ ad \pm bc }{ bd }\]

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Go ahead and see what you get
So then would the denominator be \[(1-1+x)/(1+x)\]
wait, the first part should be 1-1-x I think.
\[\huge \lim_{x \rightarrow 0} \frac{ x }{ \left( \frac{ 1-1(1+x) }{ 1+x } \right) } = \lim_{x \rightarrow 0}\frac{ x }{ \left( \frac{ 1-1-x }{ 1+x } \right) }\] correct?
Keep simplifying :)
Yeah. And then I just use the reciprocal of the denominator and plug in the 0, and I get -1?
Bingo!
thank you!!
Yw :)

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