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    AS t→infinity, x tends to a limiting value, find this limiting value
    the equation is (2e^t -2)/(2e^t -1)
    can someone explain to me what this means
    thanks
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    (Original post by mlyke)
    AS t→infinity, x tends to a limiting value, find this limiting value
    the equation is (2e^0.5 -2)/(2e^0.5 -1)
    can someone explain to me what this means
    thanks
    That's not an equation, that's a number....
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    (Original post by RDKGames)
    That's not an equation, that's a number....
    oh oops,
    replace the 0.5 with ts
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    (Original post by mlyke)
    AS t→infinity, x tends to a limiting value, find this limiting value
    the equation is (2e^t -2)/(2e^t -1)
    can someone explain to me what this means
    thanks
    Note that \displaystyle \frac{2e^t - 2}{2e^t -1} = \frac {2e^t -1 - 1}{2e^t - 1} = 1 - \frac{1}{2e^t-1}
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    (Original post by Mr M)
    Note that \displaystyle \frac{2e^t - 2}{2e^t -1} = \frac {2e^t -1 - 1}{2e^t - 1} = 1 - \frac{1}{2e^t-1}
    could you explain that, haven't learn that in class and would like to understand it
    thanks
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    (Original post by mlyke)
    could you explain that, haven't learn that in class and would like to understand it
    thanks
    The middle bit just says -2 = -1 -1 so I'm sure you follow that.

    The bit on the right just uses the fact that a number divided by itself = 1 (this is true for all numbers except 0).
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    (Original post by Mr M)
    The middle bit just says -2 = a-1 -1 so I'm sure you follow that.

    The bit on the right just uses the fact that a number divided by itself = 1 (this is true for all numbers except 0).
    Mh i understand that but how does infinity play a role here, sorry if this sounds a bit ignorant just want to understand the thought process
    The answer is 1 so its obviously right
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    (Original post by mlyke)
    oh oops,
    replace the 0.5 with ts
    Okay so you have \displaystyle x=\frac{2e^t-2}{2e^t-1}. Now you can do what Mr M has suggested. Otherwise, an alternative is to divide top and bottom of the fraction by e^t so it becomes x=\frac{2-2e^{-t}}{2-e^{-t}} and as t\rightarrow \infty you have e^{-t} \rightarrow 0 then what does x tends to?
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    (Original post by mlyke)
    Mh i understand that but how does infinity play a role here, sorry if this sounds a bit ignorant just want to understand the thought process
    The answer is 1 so its obviously right
    You end up with 1-\frac{1}{bignumber}

    Use your calculator to find the reciprocal of a really big number.
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    (Original post by Mr M)
    You end up with 1-\frac{1}{bignumber}

    Use your calculator to find the reciprocal of a really big number.


    ohhh i understand now thank you very much
 
 
 
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