In the neighbourhood of `x=0` it is known that `1+|x|lt(e^(x)-1)/(x)lt1-|x|"then find"lim_(xto0)(e^(x)-1)/(x).`

In the neighbourhood of `x=0` it is known that `1+|x|lt(e^(x)-1)/(x)lt

1.	In the neighbourhood of `x=0` it is known that `1+\|x\|lt(e^(x)-1)/(x)lt1-\|x\|"then find"lim_(xto0)(e^(x)-1)/(x).`
Answer» We have `1+\|x\|lt(e^(x)1-)/(x)lt1-\|x\|"for "x in(0-delta,0+delta).` `:." "`Using Sandwich rule, `underset(xto0)lim(1+\|x\|)ltunderset(xto0)lim(e^(x)-1)/(x)ltunderset(xto0)lim(1-\|x\|)` `implies" "1ltunderset(xto0)lim(e^(x)-1)/(x)lt1` `underset(xto0)lim(e^(x)-1)/(x)=1`

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