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Limit differential calculus

lim x approaches 0 x-ln(1+x)/1+x-e^x
Please help solve me this
Original post by Quirky01
lim x approaches 0 x-ln(1+x)/1+x-e^x
Please help solve me this


You mean this? x−ln(1+x)1+x−ex\frac{x-ln(1+x)}{1+x-e^x}

If so, use Maclaurin's expansion on ln(1+x)ln(1+x) and exe^x so you can replace them in the equation.
(edited 7 years ago)
Reply 2
Avatar for Zacken
Zacken
22
Or apply L'Hopital: lim⁡x→01−11+x1−ex=−lim⁡x→0(1+x)−2ex\displaystyle \lim_{x \to 0} \frac{1 - \frac{1}{1+x}}{1 - e^x} =- \lim_{x \to 0} \frac{(1+x)^{-2}}{e^x}
(edited 7 years ago)
Reply 3
Avatar for Quirky01
Quirky01
OP
2
Original post by Zacken
Or apply L'Hopital: lim⁡x→01−11+x1−ex=−lim⁡x→0(1+x)−2ex\displaystyle \lim_{x \to 0} \frac{1 - \frac{1}{1+x}}{1 - e^x} =- \lim_{x \to 0} \frac{(1+x)^{-2}}{e^x}


Double differentiation?
Reply 4
Avatar for Zacken
Zacken
22
Original post by Quirky01
Double differentiation?


Yep.

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