anonymous
  • anonymous
[(e^x)-(1)-(x)]/ (x^2) as lim x --> 0
Mathematics
chestercat
  • chestercat
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anonymous
  • anonymous
Again apply L' Hospital's Rule to get 1/2 as the limit.
anonymous
  • anonymous
how? =(
anonymous
  • anonymous
expand e^x and see it fall all in place ans is 1/2

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anonymous
  • anonymous
cant get the answer but i believe you
anonymous
  • anonymous
lol
anonymous
  • anonymous
how exactly did u expand e to x
anonymous
  • anonymous
The given limit is in its indeterminate form thus we differentiate the numerator and denominator to get 1/2[(e^x - 1)/x] now the limit of [e^x - 1/]x is nothing but 1.
anonymous
  • anonymous
e^x is expanded by using the Taylor Series.
anonymous
  • anonymous
That is euler series not Taylor
anonymous
  • anonymous
i didnt know you can just differentiate the numerator and the denominator while the numerator was being subtracted from e^x - 1
anonymous
  • anonymous
It is actually Maclaurin series which is a special case of Taylor Series.
anonymous
  • anonymous
If you want you can actually prove it using cauchy mean valuue theorem
anonymous
  • anonymous
can someone quickly show me how you get the answer to my other problem under this? thank you
anonymous
  • anonymous
WHere is the other problem

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