Results 1 to 5 of 5

Math Help - Limits proof

  1. #1
    Newbie
    Joined
    Sep 2009
    Posts
    5

    Limits proof

    The image attached to this post is my problem.

    My logic is that is x^2 is in the denominator, and the limit =1, then if you just take the square root of x^2 you get x, so the limit is still 1. However, I am 99% sure I am going about this wrong.

    Edit: can't change the topic name, too late now. it's a bit misleading though, sorry about that.
    Attached Thumbnails Attached Thumbnails Limits proof-screen-shot-2009-09-20-11.40.26-pm.png  
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member redsoxfan325's Avatar
    Joined
    Feb 2009
    From
    Swampscott, MA
    Posts
    943
    Quote Originally Posted by xezial View Post
    The image attached to this post is my problem.

    My logic is that is x^2 is in the denominator, and the limit =1, then if you just take the square root of x^2 you get x, so the limit is still 1. However, I am 99% sure I am going about this wrong.

    Edit: can't change the topic name, too late now. it's a bit misleading though, sorry about that.
    <br />
\lim_{x\to0}\frac{f(x)}{x}=\lim_{x\to0}\frac{xf(x)  }{x^2}=\left(\lim_{x\to0}x\right)\left(\lim_{x\to0  }\frac{f(x)}{x^2}\right)=(0)(1)=0
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Grand Panjandrum
    Joined
    Nov 2005
    From
    someplace
    Posts
    14,972
    Thanks
    4
    Quote Originally Posted by xezial View Post
    The image attached to this post is my problem.

    My logic is that is x^2 is in the denominator, and the limit =1, then if you just take the square root of x^2 you get x, so the limit is still 1. However, I am 99% sure I am going about this wrong.

    Edit: can't change the topic name, too late now. it's a bit misleading though, sorry about that.
    We are given:

    \lim_{x \to 0} \frac{f(x)}{x^2}=1


    So for any \varepsilon>0, for x small enough:

    1-\varepsilon<\frac{f(x)}{x^2}<1+\varepsilon

    or:

    (1-\varepsilon)x<\frac{f(x)}{x}<(1+\varepsilon)x

    hence by the squeeze theorem \lim_{x \to 0} \frac{f(x)}{x}=??

    CB
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Newbie
    Joined
    Sep 2009
    Posts
    5
    Wow, I haven't really gotten that far into calc to be familiar with the squeeze theorem and all that... but is the solution redsox fan posted legitamate?
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Flow Master
    mr fantastic's Avatar
    Joined
    Dec 2007
    From
    Zeitgeist
    Posts
    16,948
    Thanks
    5
    Quote Originally Posted by xezial View Post
    Wow, I haven't really gotten that far into calc to be familiar with the squeeze theorem and all that... but is the solution redsox fan posted legitamate?
    Yes.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. limits proof
    Posted in the Calculus Forum
    Replies: 7
    Last Post: December 23rd 2010, 05:13 AM
  2. Limits Proof
    Posted in the Calculus Forum
    Replies: 1
    Last Post: September 10th 2009, 12:49 AM
  3. Proof with limits
    Posted in the Calculus Forum
    Replies: 2
    Last Post: October 30th 2008, 01:15 PM
  4. proof of 2 limits
    Posted in the Calculus Forum
    Replies: 2
    Last Post: September 25th 2007, 10:55 AM
  5. limits/proof
    Posted in the Calculus Forum
    Replies: 4
    Last Post: April 8th 2007, 03:44 PM

Search Tags


/mathhelpforum @mathhelpforum