Results 1 to 11 of 11

Math Help - L'Hospital's Rule

  1. #1
    Member mybrohshi5's Avatar
    Joined
    Sep 2009
    Posts
    230

    Angry L'Hospital's Rule

    Use L'Hospital's Rule to find the following limit:

     \lim_{x\rightarrow\infty} (1+\frac {2}{x})^{x}

    I am not sure how to use L'Hospital's rule for this.

    I know that the  \lim_{x\rightarrow\infty} \frac{2}{x} = 0

    and so therefore,

     \lim_{x\rightarrow\infty} (1 + 0)^{x} = 1^{\infty} = 1

    Can someone help me solve this using L'Hospital's Rule.

    Thank you.
    Last edited by mybrohshi5; December 9th 2009 at 02:56 PM.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member
    Joined
    Jun 2009
    From
    Africa
    Posts
    641

    Smile

    Quote Originally Posted by mybrohshi5 View Post
    Use L'Hospital's Rule to find the following limit:

     \lim_{x\rightarrow\infty} (1+\frac {2}{x})^{x}

    I am not sure how to use L'Hospital's rule for this.

    I know that the  \lim_{x\rightarrow\infty} \frac{2}{x} = 0

    and so therefore,

     \lim_{x\rightarrow\infty} (1 + 0)^{x} = 1^{\infty} = 1

    Can someone help me solve this using L'Hospital's Rule.

    Thank you.
    \lim_{x\to \infty }\left (1+\frac{2}{x}  \right ) =\lim_{x\to \infty }\left (\frac{x+2}{x}  \right ) =\lim_{x\to \infty }(\frac{1}{1})=1 (by the Hospital's rule).
    therefore \lim_{x\to \infty }(1)^x=1
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Super Member
    Joined
    Aug 2009
    From
    Israel
    Posts
    976
    That's wrong, Raoh :P

    \lim_{x\to \infty }\left (1+\frac{1}{x} \right ) = \lim_{x\to \infty }\left (\frac{x+1}{x} \right ) = 1, but
    \lim_{x\to \infty }\left (1+\frac{1}{x} \right )^x = e

    What you need to do here:
    Set t = \left (1+\frac{2}{x} \right )^x, evaluate ln(t) and take the limit as x goes to infinity. (Hint: the answer is not 1)
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Super Member
    Joined
    Jun 2009
    From
    Africa
    Posts
    641

    Smile

    Quote Originally Posted by Defunkt View Post
    That's wrong, Raoh :P

    \lim_{x\to \infty }\left (1+\frac{1}{x} \right ) = \lim_{x\to \infty }\left (\frac{x+1}{x} \right ) = 1, but
    \lim_{x\to \infty }\left (1+\frac{1}{x} \right )^x = e

    What you need to do here:
    Set t = \left (1+\frac{2}{x} \right )^x, evaluate ln(t) and take the limit as x goes to infinity. (Hint: the answer is not 1)
    errr !
    \ln(t)=x\ln(1+\frac{2}{x})
    \lim_{x\to \infty }\left (x\ln(1+\frac{2}{x})  \right )=2.
    hope it's right now
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Member mybrohshi5's Avatar
    Joined
    Sep 2009
    Posts
    230
    Im not exactly sure how by setting t equal to the function then taking the natural log of it is using L'Hospitals Rule?

    Could you maybe show me in a little more detail what you mean.

    Thank you
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Member mybrohshi5's Avatar
    Joined
    Sep 2009
    Posts
    230
    Quote Originally Posted by Raoh View Post
    errr !
    \ln(t)=x\ln(1+\frac{2}{x})
    \lim_{x\to \infty }\left (x\ln(1+\frac{2}{x})  \right )=2.
    hope it's right now
    Could you show me how \lim_{x\to \infty }\left (x\ln(1+\frac{2}{x})  \right )=2

    Im not really understanding how it equals two
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Super Member
    Joined
    Jun 2009
    From
    Africa
    Posts
    641

    Smile

    Quote Originally Posted by mybrohshi5 View Post
    Could you show me how \lim_{x\to \infty }\left (x\ln(1+\frac{2}{x})  \right )=2

    Im not really understanding how it equals two
    put x=\frac{1}{t} when x\to \infty we will have t\to 0^+.
    therefore we should evaluate,
    \lim_{t\to 0^+}
    now use the hospital's rule.
    you'll get,

    hope that right.
    Last edited by Raoh; December 9th 2009 at 02:33 PM.
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Member mybrohshi5's Avatar
    Joined
    Sep 2009
    Posts
    230
    When i graph the equation it looks like it is approaching about 7.4 when x goes to infinity????
    Follow Math Help Forum on Facebook and Google+

  9. #9
    MHF Contributor
    skeeter's Avatar
    Joined
    Jun 2008
    From
    North Texas
    Posts
    11,690
    Thanks
    448
    Quote Originally Posted by mybrohshi5 View Post
    When i graph the equation it looks like it is approaching about 7.4 when x goes to infinity????
    that's because the limit is e^2
    Follow Math Help Forum on Facebook and Google+

  10. #10
    Member mybrohshi5's Avatar
    Joined
    Sep 2009
    Posts
    230
    Could you please show me how to use L'Hospital's Rule to show that it is e^2 cause Naoh tried it and came up with 2
    Follow Math Help Forum on Facebook and Google+

  11. #11
    Super Member
    Joined
    Aug 2009
    From
    Israel
    Posts
    976
    He came up with \lim_{x\to \infty }ln(t)= \lim_{x\to \infty }\left (x\ln(1+\frac{2}{x}) \right )=2

    Now use the fact that e^{lnx} = x to conclude the result.. (remember - you're looking for t)
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. without using L'Hospital's Rule
    Posted in the Calculus Forum
    Replies: 1
    Last Post: December 31st 2010, 01:48 AM
  2. L'Hospital's rule
    Posted in the Calculus Forum
    Replies: 8
    Last Post: February 16th 2010, 07:37 PM
  3. L'Hospital's rule
    Posted in the Calculus Forum
    Replies: 10
    Last Post: November 19th 2009, 07:31 PM
  4. L'Hospital's Rule
    Posted in the Calculus Forum
    Replies: 3
    Last Post: November 15th 2009, 12:57 PM
  5. L'Hospital's Rule
    Posted in the Calculus Forum
    Replies: 2
    Last Post: October 15th 2008, 07:24 PM

Search Tags


/mathhelpforum @mathhelpforum