Zenmo
  • Zenmo
Find the limit if it exists. Use L' Hospital rule.
Mathematics
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SOLVED
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jamiebookeater
  • jamiebookeater
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Zenmo
  • Zenmo
\[\lim_{x \rightarrow 0}(\frac{ 1 }{ x }-\frac{ 1 }{ e^x-1 })\]
jim_thompson5910
  • jim_thompson5910
what do you get when you combine the fractions?
Zenmo
  • Zenmo
\[\lim_{x \rightarrow 0}(\frac{ (e^x-1 )-x}{ x(e^x-1) }\]

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jim_thompson5910
  • jim_thompson5910
I'm getting the same
jim_thompson5910
  • jim_thompson5910
if you plug in x = 0, you should get 0/0 which is one of the many indeterminate forms so use L'Hospital's rule to get what?
Zenmo
  • Zenmo
do I simplify the bottom first then use the L'hospital rule?
jim_thompson5910
  • jim_thompson5910
simplify how?
Zenmo
  • Zenmo
\[xe^x-x\]
jim_thompson5910
  • jim_thompson5910
sure you can distribute
Zenmo
  • Zenmo
\[\lim_{x \rightarrow 0}\frac{ (e^x-1)-1 }{ xe^x \times e^x-1 }\]
Zenmo
  • Zenmo
then use L'Hospital rule again?
jim_thompson5910
  • jim_thompson5910
after using L'Hospital rule the first time, you should have \[\Large \lim_{x \to 0}\frac{e^x-1}{e^x+xe^x-1}\]
jim_thompson5910
  • jim_thompson5910
plugging in x = 0 into that will lead to 0/0 again so you have to use L'Hospital's rule again
Zenmo
  • Zenmo
but isn't \[(e^x-1)-x = e^x-1\] since (-x) = -1?
Zenmo
  • Zenmo
\[(e^x-1)-1\]**
jim_thompson5910
  • jim_thompson5910
taking the derivative of -1 is 0
Zenmo
  • Zenmo
oh you did the L'hospital rule the 2nd time then.
jim_thompson5910
  • jim_thompson5910
Original \[\Large \lim_{x \to 0}\frac{ (e^x-1 )-x}{ x(e^x-1) }\] -------------------------------------------------- after applying L'Hospital's rule the first time, you'll get \[\Large \lim_{x \to 0}\frac{e^x-1}{e^x+xe^x-1}\] -------------------------------------------------- after applying L'Hospital's rule the second time, you'll get \[\Large \lim_{x \to 0}\frac{e^x}{2e^x+xe^x}\]
Zenmo
  • Zenmo
then since top and bottom isn't 0/0, we can plug in 0 to get 1/2?
jim_thompson5910
  • jim_thompson5910
1/2 is your final answer, yep
Zenmo
  • Zenmo
okay, got it. Thanks! :D
jim_thompson5910
  • jim_thompson5910
np

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