Results 1 to 6 of 6

Math Help - Limit of (a^x -1)/x when x approaches 0

  1. #1
    Newbie
    Joined
    Mar 2010
    Posts
    15

    Limit of (a^x -1)/x when x approaches 0

    The answer is ln a.

    This was a problem we got last semester (book of exercises). We didn't have to solve this but I'm still wondering how to solve it.
    The tip is to substitute a^x - 1 by t.
    I would enter the formula in LaTeX but I haven't found yet how to enter a limit in LaTeX-code.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Member Miss's Avatar
    Joined
    Mar 2010
    From
    Planet earth.
    Posts
    165
    Quote Originally Posted by Bart View Post
    The answer is ln a.

    This was a problem we got last semester (book of exercises). We didn't have to solve this but I'm still wondering how to solve it.
    The tip is to substitute a^x - 1 by t.
    I would enter the formula in LaTeX but I haven't found yet how to enter a limit in LaTeX-code.
    Let f(x)=a^x , then f(0)=1.

    \lim_{x\to 0} \frac{a^x-1}{x}=\lim_{x\to 0} \frac{a^x-f(0)}{x-0}=f'(0)=a^x ln(a) |_{x=0}=ln(a)
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor
    Joined
    Dec 2009
    Posts
    3,120
    Thanks
    1
    Quote Originally Posted by Bart View Post
    The answer is ln a.

    This was a problem we got last semester (book of exercises). We didn't have to solve this but I'm still wondering how to solve it.
    The tip is to substitute a^x - 1 by t.
    I would enter the formula in LaTeX but I haven't found yet how to enter a limit in LaTeX-code.
    Alternatively, in terms of you substitution

    \lim_{x\to 0}\left(\frac{a^x-1}{x}\right)

    a^x-1=t,\ a^x=t+1

    x=0\ \Rightarrow\ 1=t+1\ \Rightarrow\ t=0

    lna^x=ln(t+1)

    xlna=ln(t+1)

    x=\frac{ln(t+1)}{lna}

    \frac{1}{x}=\frac{lna}{ln(t+1)}

    \lim_{x\to 0}\left(\frac{a^x-1}{x}\right)=\left(lna\right)\lim_{t\to 0}\left(\frac{t}{ln(t+1)}\right)

    Finally use

    \lim_{t\to 0}\left[ln(t+1)\right]=t
    Last edited by Archie Meade; March 23rd 2010 at 10:29 AM. Reason: typo
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Newbie
    Joined
    Mar 2010
    Posts
    15
    Quote Originally Posted by Archie Meade View Post
    Alternatively, in terms of you substitution

    \lim_{x\to 0}\left(\frac{a^x-1}{x}\right)

    a^x-1=t,\ a^x=t+1

    x=0\ \Rightarrow\ 1=t+1\ \Rightarrow\ t=0

    lna^x=ln(t+1)

    xlna=ln(t+1)

    x=\frac{ln(t+1)}{lna}

    \frac{1}{x}=\frac{lna}{ln(t+1)}

    \lim_{x\to 0}\left(\frac{a^x-1}{x}\right)=\left(lna\right)\lim_{t\to 0}\left(\frac{t}{t+1}\right)

    Finally use

    \lim_{t\to 0}\left[ln(t+1)\right]=t
    Thank you both, this was the solution I needed.
    I forgot to mention that we had to solve all limits without l' H˘spital, just for the algebraic exercise.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    MHF Contributor
    Joined
    Dec 2009
    Posts
    3,120
    Thanks
    1
    Hi Bart,

    I had a typo on the 2nd last line.

    You could finally use the series expansion for ln(1+x)

    ln(1+x)=x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+.......

    As x approaches zero, the higher powers of x become increasingly insignificant
    in comparison to x,

    hence

    ln(1+t)\ \rightarrow\ t,\ as\ t\ \rightarrow\ 0
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Newbie
    Joined
    Mar 2010
    Posts
    15
    I thought so.
    Thanks.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. A limit as x approaches a.
    Posted in the Calculus Forum
    Replies: 14
    Last Post: February 11th 2011, 04:03 AM
  2. Limit as x approaches 0
    Posted in the Calculus Forum
    Replies: 5
    Last Post: April 15th 2010, 07:41 AM
  3. when x approaches to -inf in limit
    Posted in the Calculus Forum
    Replies: 5
    Last Post: February 1st 2010, 07:01 AM
  4. the limit as x approaches infinity
    Posted in the Calculus Forum
    Replies: 3
    Last Post: September 15th 2008, 01:06 PM
  5. limit as h approaches 0
    Posted in the Calculus Forum
    Replies: 4
    Last Post: January 13th 2008, 09:50 AM

Search Tags


/mathhelpforum @mathhelpforum