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Math Help - limit..again

  1. #1
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    limit..again

    allright sorry for asking a million questions..but this one is weird also.
    the limit as x goes to 0 ((e^x) - 1) times cosx
    the answer in the book is 1 but I always get an undefined answer when I put that limit into a fraction. (either the cosx on the bottom of the fraction..((e^x) - 1)^-1/cosx..or secx/e^x - 1 both give me invalid answers.) Thx for the help.


    edit..ok I looked online for the mistakes in book. It should be ((e^x-1) times cotx
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  2. #2
    MHF Contributor red_dog's Avatar
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    \displaystyle \lim_{x\to 0}(e^x-1)\cot x=\lim_{x\to 0}\frac{e^x-1}{\tan x}=\lim_{x\to 0}\frac{e^x-1}{x}\cdot\frac{x}{\tan x}=1\cdot 1=1
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  3. #3
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    Quote Originally Posted by red_dog View Post
    \displaystyle \lim_{x\to 0}(e^x-1)\cot x=\lim_{x\to 0}\frac{e^x-1}{\tan x}=\lim_{x\to 0}\frac{e^x-1}{x}\cdot\frac{x}{\tan x}=1\cdot 1=1

    I saw you broke that up into two fractions. How does that work and why did you do that?
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  4. #4
    MHF Contributor red_dog's Avatar
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    If \displaystyle \lim_{x\to x_0}f(x)=l_1 and \displaystyle \lim_{x\to x_0}g(x)=l_2 then \displaystyle \lim_{x\to x_0}f(x)g(x)=\displaystyle \lim_{x\to x_0}f(x)\cdot \displaystyle \lim_{x\to x_0}g(x)=l_1\cdot l_2. (Exception 0\cdot\infty).

    \displaystyle \lim_{x\to 0}\frac{e^x-1}{x}=1 by l'Hospital
    and \displaystyle \lim_{x\to 0}\frac{x}{\tan x}=1 is a remarcable limit.
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