anonymous
  • anonymous
can someone help me evaluate the limit of this equation limit approaches 0(e^(4x) - 1-4x)/(x^2)
Mathematics
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anonymous
  • anonymous
can someone help me evaluate the limit of this equation limit approaches 0(e^(4x) - 1-4x)/(x^2)
Mathematics
jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
0
anonymous
  • anonymous
i get zero over zero
amistre64
  • amistre64
derive top and bottom seperately; then limit them seperately

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amistre64
  • amistre64
lhopitals rule
amistre64
  • amistre64
i wonder if you can do that more than once
anonymous
  • anonymous
Yes. You can do it as long as both the denominator and numerator have 0/0 or ∞/∞
amistre64
  • amistre64
2x doesnt seem to help does it lol
amistre64
  • amistre64
you can? over and over?
anonymous
  • anonymous
yeah as long as you get 0/0 or infin/infin
amistre64
  • amistre64
i thought i read that somewhere lol
anonymous
  • anonymous
ok so i would get the derivative of the top correct over the derivative of the bottom
amistre64
  • amistre64
yes
amistre64
  • amistre64
4.e^4x - 4 2.e^4x - 2 ---------- = --------- 2x x
amistre64
  • amistre64
8.e^4x at 0 = ...8 right?
anonymous
  • anonymous
yeah thats what the answer sheet got but i still dont understand how you got it
amistre64
  • amistre64
x^2; 2x ; 2 tese are the derivatives right? of the bottom
anonymous
  • anonymous
yeah
amistre64
  • amistre64
e^4x -1 - 4x; 4.e^4x -4; 16.e^4x for the top
amistre64
  • amistre64
16.e^4x ------- = 8.e^4x ; at x=0; 8.1 = 8 2
anonymous
  • anonymous
cant see where you get the 16 from sorry
amistre64
  • amistre64
4.e^4x right? derive e^4x again and we get: 4. 4.e^4x = 16.e^4x
amistre64
  • amistre64
e^u derives to : Du e^u
anonymous
  • anonymous
ok so u used l hopital twice
amistre64
  • amistre64
yes lol
anonymous
  • anonymous
ok ok awesome and we were just talking about that

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