anonymous
  • anonymous
lim(xto0)2[(e^x-(1+x))/x^2]=
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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anonymous
  • anonymous
\[\lim_{x\to 0} e^x-\frac{(1+x)}{x^2}\] ?
anonymous
  • anonymous
satellite u lost 2
anonymous
  • anonymous
ooh i know \[lim_{x\to 0} \frac{2e^x-(1+x)}{x^2}\]

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anonymous
  • anonymous
i bet that is it. l'hopital's rule problem. of the form \[\frac{0}{0}\]
anonymous
  • anonymous
take derivative top and bottom separate. you will need to do it twice because after the first time you still get \[\frac{0}{0}\]
myininaya
  • myininaya
the top is 2e^0-1-0=2(1)-1-0=2-1=1
anonymous
  • anonymous
first time \[\frac{2e^x-1}{2x}\]
anonymous
  • anonymous
still \[\frac{0}{0}\]
anonymous
  • anonymous
second time \[\frac{2e^x}{2}\] let x ->0 get 1
anonymous
  • anonymous
oops i am totally wrong. sorry. first time it is \[\frac{1}{0}\] no limit. so sorry.
anonymous
  • anonymous
must be my bed time.
anonymous
  • anonymous
yes you have lost the derivative of constant 1 first time
anonymous
  • anonymous
by first time i mean before you start. it is not \[\frac{0}{0}\] to begin with so no l'hopital. sorry
anonymous
  • anonymous
WHERE IS THE DERIVATIVE OF CONSTANT?
anonymous
  • anonymous
Mmmm....looking carefully at the question as posted, we are looking for the limit of 2(e^x - 1 - x)/x^2. As x goes to zero we get 2(e^0 - 1 - 0)/0^2 = 2(1 - 1)/0 = 0/0, so use l'Hopital twice to get Lim as x goes to 0 of e^x, which is 1.
myininaya
  • myininaya
did you try graphing it? both sides of 0 f(x) goes to positive infinity
anonymous
  • anonymous
WHY WE TAKE DERIVATIVE ANY REASON????

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