Facebook Page
Twitter
RSS
+ Reply to Thread
Results 1 to 4 of 4

Thread: limit

  1. MHB Craftsman

    Status
    Offline
    Join Date
    Jan 2014
    Posts
    236
    Thanks
    133 times
    Thanked
    41 times
    #1
    $\lim_{{x}\to{0}} \frac{3^x - 2^x}{x^2 - x}$

    I'm trying to solve this limit. I'm not sure how to go about it. I can try multiplying the numerator and denominator by either $\frac{1}{x^2}$, $\frac{1}{x}$, $\frac{1}{3^x}$, $\frac{1}{2^x}$, but that doesn't lead anywhere.

  2. Pessimist Singularitarian
    MHB Coder
    MHB Math Helper
    MHB Ambassador
    MarkFL's Avatar
    Status
    Offline
    Join Date
    Feb 2012
    Location
    St. Augustine, FL.
    Posts
    11,982
    Thanks
    32,421 times
    Thanked
    28,357 times
    Thank/Post
    2.367
    Trophies
    24 Highscores
    Awards
    MHB Statistics Award (2016)  

MHB Calculus Award (2016)  

MHB Pre-University Math Award (2016)  

MHB Model Helper Award (2015)  

MHB Calculus Award (2015)
    #2
    We are given to evaluate:

    $ \displaystyle L=\lim_{x\to0}\left(\frac{3^x-2^x}{x^2-x}\right)$

    We have the indeterminate form 0/0, so we may apply L'H˘pital's Rule:

    $ \displaystyle L=\lim_{x\to0}\left(\frac{\ln(3)3^x-\ln(2)2^x}{2x-1}\right)=$?

  3. Perseverance
    MHB Global Moderator
    MHB Math Helper
    greg1313's Avatar
    Status
    Offline
    Join Date
    Feb 2013
    Location
    London, Ontario, Canada - The Forest City
    Posts
    996
    Thanks
    3,897 times
    Thanked
    1,996 time
    Thank/Post
    2.004
    #3
    What about series? You've posted in Pre-Calculus, so I'm guessing neither of these methods are available.

  4. Pessimist Singularitarian
    MHB Coder
    MHB Math Helper
    MHB Ambassador
    MarkFL's Avatar
    Status
    Offline
    Join Date
    Feb 2012
    Location
    St. Augustine, FL.
    Posts
    11,982
    Thanks
    32,421 times
    Thanked
    28,357 times
    Thank/Post
    2.367
    Trophies
    24 Highscores
    Awards
    MHB Statistics Award (2016)  

MHB Calculus Award (2016)  

MHB Pre-University Math Award (2016)  

MHB Model Helper Award (2015)  

MHB Calculus Award (2015)
    #4
    Oops...I didn't notice this wasn't posted in the Calculus forum. So L'H˘pital is out.

    Okay, let's consider:

    $ \displaystyle \frac{3^x-2^x}{x^2-x}=\frac{2^x}{x-1}\cdot\frac{\left(\dfrac{3}{2}\right)^x-1}{x}$

    Next, let's consider the limit:

    $ \displaystyle L_a=\lim_{x\to0}\left(\frac{a^x-1}{x}\right)$ where $0<a$

    Let:

    $ \displaystyle a^x=y+1\implies x=\frac{\ln(y+1)}{\ln(a)}$

    And we observe that as $x\to0$ we have $y\to0$, hence:

    $ \displaystyle L_a=\lim_{y\to0}\left(\frac{\ln(a)}{\ln\left((y+1)^{\frac{1}{y}}\right)}\right)$

    Now, using $ \displaystyle \lim_{u\to0}\left((1+u)^{\frac{1}{u}}\right)=e$ we conclude:

    $ \displaystyle L_a=\ln(a)$

    And so:

    $ \displaystyle L=\lim_{x\to0}\left(\frac{2^x}{x-1}\right)\cdot\lim_{x\to0}\left(\frac{\left(\dfrac{3}{2}\right)^x-1}{x}\right)=(-1)\left(\ln\left(\frac{3}{2}\right)\right)=\ln\left(\frac{2}{3}\right)$

Similar Threads

  1. [SOLVED] Limit of lnx m* x?
    By tmt in forum Pre-Calculus
    Replies: 1
    Last Post: December 4th, 2016, 20:58
  2. [SOLVED] Evaluating limit limit of multivariable function
    By tmt in forum Calculus
    Replies: 3
    Last Post: July 31st, 2016, 18:48
  3. [SOLVED] Limit
    By Guest in forum Calculus
    Replies: 1
    Last Post: January 4th, 2016, 08:23
  4. limit superior and limit inferior
    By suvadip in forum Analysis
    Replies: 1
    Last Post: February 21st, 2014, 08:15
  5. Limit is Sum
    By jeffer vitola in forum Challenge Questions and Puzzles
    Replies: 4
    Last Post: August 18th, 2013, 03:10

Tags for this Thread

Posting Permissions

  • You may not post new threads
  • You may not post replies
  • You may not post attachments
  • You may not edit your posts
  •  
Math Help Boards