Results 1 to 9 of 9

Math Help - calculus - limits

  1. #1
    Newbie
    Joined
    Mar 2008
    Posts
    9

    calculus - limits

    evaluate:

    lim x approach infinity [sqrt(e^2x-e^x)-e^x]

    tried alotta things, didn work, help pls!
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Flow Master
    mr fantastic's Avatar
    Joined
    Dec 2007
    From
    Zeitgeist
    Posts
    16,948
    Thanks
    5
    Quote Originally Posted by ssdimensionss View Post
    evaluate:

    lim x approach infinity [sqrt(e^2x-e^x)-e^x]

    tried alotta things, didn work, help pls!
    = \lim_{x \rightarrow + \infty}\frac{\sqrt{(e^{2x} - e^x} - e^x)(\sqrt{e^{2x} - e^x} + e^x)}{\sqrt{e^{2x} - e^x} + e^x}


    = \lim_{x \rightarrow + \infty}\frac{e^{2x} - e^x - e^{2x}}{\sqrt{e^{2x} - e^x} + e^x}


    = \lim_{x \rightarrow + \infty} - \frac{e^x}{\sqrt{e^{2x} - e^x} + e^x}


    = \lim_{x \rightarrow + \infty} - \frac{1}{\sqrt{1 - e^{-x}} + 1}


    = - \frac{1}{2}.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Mar 2008
    Posts
    9

    thanks

    hey thx for the help! i did it all up to the second to last step, the bit i dont get is why does e^x limit for x apporach infinty = 1?? please clarify, thx again!
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Flow Master
    mr fantastic's Avatar
    Joined
    Dec 2007
    From
    Zeitgeist
    Posts
    16,948
    Thanks
    5
    Quote Originally Posted by ssdimensionss View Post
    hey thx for the help! i did it all up to the second to last step, the bit i dont get is why does e^x limit for x apporach infinty = 1?? please clarify, thx again!
    e^{-x} \rightarrow 0 as x \rightarrow +\infty. So you could insert the following line between the second last and last:

    = - \frac{1}{\sqrt{1 - 0} + 1}

    Capisce?
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Newbie
    Joined
    Mar 2008
    Posts
    9

    ^

    yes but that still does not explain why the numerator which was lim x approach infinity e^x is = 1. sorry for bothering again!
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Moo
    Moo is offline
    A Cute Angle Moo's Avatar
    Joined
    Mar 2008
    From
    P(I'm here)=1/3, P(I'm there)=t+1/3
    Posts
    5,618
    Thanks
    6
    Hello,

    He simplified the numerator and the denominator with e^x, so e^x becomes 1.
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Flow Master
    mr fantastic's Avatar
    Joined
    Dec 2007
    From
    Zeitgeist
    Posts
    16,948
    Thanks
    5
    Quote Originally Posted by ssdimensionss View Post
    yes but that still does not explain why the numerator which was lim x approach infinity e^x is = 1. sorry for bothering again!
    I divided the numerator and denominator by e^x. When you divide the numerator by e^x you get 1.

    Note on denominator: When dividing outside the square root by e^x, you divide inside the square root by e^{2x}:

    \frac{\sqrt{e^{2x} - e^x}}{e^x} = \sqrt{\frac{e^{2x} - e^x}{e^{2x}}} = \sqrt{1 - e^{-x}}.
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Newbie
    Joined
    Mar 2008
    Posts
    9
    i understand now thx so much!
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Newbie
    Joined
    Mar 2008
    Posts
    9
    i understand now thx so much!
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Calculus 2 limits as x-->0+
    Posted in the Calculus Forum
    Replies: 3
    Last Post: August 3rd 2010, 06:28 PM
  2. Calculus, limits
    Posted in the Calculus Forum
    Replies: 2
    Last Post: April 19th 2010, 05:03 PM
  3. Calculus, limits
    Posted in the Calculus Forum
    Replies: 10
    Last Post: January 31st 2010, 06:17 PM
  4. Calculus Limits Help
    Posted in the Calculus Forum
    Replies: 2
    Last Post: January 31st 2009, 01:22 PM
  5. calculus limits
    Posted in the Calculus Forum
    Replies: 2
    Last Post: January 31st 2007, 03:53 AM

Search Tags


/mathhelpforum @mathhelpforum