Results 1 to 10 of 10

Math Help - Proving limits

  1. #1
    Junior Member
    Joined
    Oct 2010
    Posts
    71

    Proving limits

    Prove (from the definition) that if f(x) tends to L as x tends to a then 2f(x) tends to 2L as x tends to a

    Now we can suppose (For all epsilon>0) (There exists delta >0) (For all x E X)
    0<|x-a|<delta implies |f(x) - L| < epsilon.

    Now we need to prove that 2f(x) tends to 2L as x tends to a.

    I'm struggling a bit with how to do this..
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Joined
    Oct 2009
    Posts
    5,530
    Thanks
    774
    Now we can suppose (For all epsilon>0) (There exists delta >0) (For all x E X) 0<|x-a|<delta implies |f(x) - L| < epsilon.
    This means that somebody is able to return a \delta for each \varepsilon given to them. It follows that for every \varepsilon>0 there exists a \delta>0 such that for all x\in X, 0<|x-a|<\delta implies |f(x) - L| < \varepsilon/2. Indeed, given an \varepsilon, give \varepsilon/2 to the \delta-providing person that exists according to your statement.

    Can you finish?
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    Oct 2010
    Posts
    71
    Quote Originally Posted by emakarov View Post
    This means that somebody is able to return a \delta for each \varepsilon given to them. It follows that for every \varepsilon>0 there exists a \delta>0 such that for all x\in X, 0<|x-a|<\delta implies |f(x) - L| < \varepsilon/2. Indeed, given an \varepsilon, give \varepsilon/2 to the \delta-providing person that exists according to your statement.

    Can you finish?
    No, I can't. I don't quite understand your explanation ..

    I am able to do the limit questions where f(x) , the limit and what x tends to is given. I'm just confused by this particular one..
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor
    Joined
    Oct 2009
    Posts
    5,530
    Thanks
    774
    One has to show:

    for every \varepsilon>0 there exists a \delta>0 such that for all x\in X, 0<|x-a|<\delta implies |f(x) - L| < \varepsilon/2. (*)

    Since |f(x) - L| < \varepsilon/2 is equivalent to |2f(x) - 2L| < \varepsilon, (*) means that \lim\limits_{x\to a}2f(x)=2L.

    How do we show (*)? We are given

    for every \varepsilon>0 there exists a \delta>0 such that for all x\in X, 0<|x-a|<\delta implies |f(x) - L| < \varepsilon. (**)

    The latter statement means that there is a jinny that, given any \varepsilon, returns a \delta such that for every x bla, bla, bla. Now, you need to do something similar to that jinny to show (*). Suppose somebody gives you an \varepsilon. You halve it and give \varepsilon/2 to the jinny. True to its task, the jinny gives you back a \delta such that for all x\in X, 0<|x-a|<\delta implies |f(x) - L| < \varepsilon/2. You take this \delta and return it as your own.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Junior Member
    Joined
    Oct 2010
    Posts
    71
    Quote Originally Posted by emakarov View Post
    One has to show:

    for every \varepsilon>0 there exists a \delta>0 such that for all x\in X, 0<|x-a|<\delta implies |f(x) - L| < \varepsilon/2. (*)

    Since |f(x) - L| < \varepsilon/2 is equivalent to |2f(x) - 2L| < \varepsilon, (*) means that \lim\limits_{x\to a}2f(x)=2L.

    How do we show (*)? We are given

    for every \varepsilon>0 there exists a \delta>0 such that for all x\in X, 0<|x-a|<\delta implies |f(x) - L| < \varepsilon. (**)

    The latter statement means that there is a jinny that, given any \varepsilon, returns a \delta such that for every x bla, bla, bla. Now, you need to do something similar to that jinny to show (*). Suppose somebody gives you an \varepsilon. You halve it and give \varepsilon/2 to the jinny. True to its task, the jinny gives you back a \delta such that for all x\in X, 0<|x-a|<\delta implies |f(x) - L| < \varepsilon/2. You take this \delta and return it as your own.
    I understand that a little more now, thanks.

    It makes sense that I need to show (*) but I still don't get how.

    Can I say let delta = epsilon/2 ?

    So 0<|x-a| < delta implies |f(x) - l| < epsilon/2

    Then that is equaivalent to |2f(x) - 2L| < epsilon ? ? ?
    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor

    Joined
    Apr 2005
    Posts
    15,598
    Thanks
    1421
    Quote Originally Posted by chr91 View Post
    I understand that a little more now, thanks.

    It makes sense that I need to show (*) but I still don't get how.

    Can I say let delta = epsilon/2 ?
    No, you can't and no one has suggested such a thing.

    Since f(x)->L, given any \epsilon>0 there exist \delta> 0 such that if |x- a|< \delta then |f(x)- L|< \epsilon.
    That's the definition of "limit".

    What they are saying is that to prove 2f(x)-> 2L, use the \delta that the previous statement gives for \epsilon/2 rather than \epsilon.

    So 0<|x-a| < delta implies |f(x) - l| < epsilon/2

    Then that is equaivalent to |2f(x) - 2L| < epsilon ? ? ?
    That last part is correct but does NOT imply that you use \delta/2. The relationship between \epsilon and \delta is NOT in general "linear"!
    Follow Math Help Forum on Facebook and Google+

  7. #7
    MHF Contributor
    Joined
    Oct 2009
    Posts
    5,530
    Thanks
    774
    Can I say let delta = epsilon/2 ?
    No. The relationship between epsilon and delta can be very complex depending on f. Epsilon measures the variation of f(x) whereas delta measures the variation of x.

    So 0<|x-a| < delta implies |f(x) - l| < epsilon/2
    I don't see how this follows.

    The difficulty here may be in using the same variable name \varepsilon in two different statements. Let me rewrite this.

    We are given

    for every \alpha>0 there exists a \beta>0 such that for all x\in X, 0<|x-a|<\beta implies |f(x) - L| < \alpha. (**)

    We need to show

    for every \varepsilon>0 there exists a \delta>0 such that for all x\in X, 0<|x-a|<\delta implies |f(x) - L| < \varepsilon/2. (*)

    We start the proof in the standard way. Fix an arbitrary \varepsilon. Instantiate \alpha to \varepsilon/2 in (**). Then (**) says that there exists a \beta such that 0<|x-a|<\beta implies |f(x) - L| < \varepsilon/2. So, take \delta=\beta. Then for every x, if 0<|x-a|<\delta, then |f(x) - L| < \varepsilon/2, as required to prove (*).
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Junior Member
    Joined
    Oct 2010
    Posts
    71
    Quote Originally Posted by emakarov View Post
    No. The relationship between epsilon and delta can be very complex depending on f. Epsilon measures the variation of f(x) whereas delta measures the variation of x.

    I don't see how this follows.

    The difficulty here may be in using the same variable name \varepsilon in two different statements. Let me rewrite this.

    We are given

    for every \alpha>0 there exists a \beta>0 such that for all x\in X, 0<|x-a|<\beta implies |f(x) - L| < \alpha. (**)

    We need to show

    for every \varepsilon>0 there exists a \delta>0 such that for all x\in X, 0<|x-a|<\delta implies |f(x) - L| < \varepsilon/2. (*)

    We start the proof in the standard way. Fix an arbitrary \varepsilon. Instantiate \alpha to \varepsilon/2 in (**). Then (**) says that there exists a \beta such that 0<|x-a|<\beta implies |f(x) - L| < \varepsilon/2. So, take \delta=\beta. Then for every x, if 0<|x-a|<\delta, then |f(x) - L| < \varepsilon/2, as required to prove (*).
    Thanks so much for your time and effort to help. I understand it now.

    I have a similar question though. It's the same question apart from if f(x) tends to L as x tends to a, then |f(x)| tends to |L| as x tends to a..


    Is the format exactly the same but instead of epsilon / 2, we have mod of epsilon?
    Follow Math Help Forum on Facebook and Google+

  9. #9
    MHF Contributor
    Joined
    Oct 2009
    Posts
    5,530
    Thanks
    774
    Is the format exactly the same but instead of epsilon / 2, we have mod of epsilon?
    Everywhere we assumed that epsilon > 0, so |epsilon| = epsilon. Here you can instantiate alpha in (**) to just epsilon because once |f(x) - L| < epsilon, | |f(x)| - |L| | <= |f(x) - L| < epsilon.
    Follow Math Help Forum on Facebook and Google+

  10. #10
    MHF Contributor

    Joined
    Apr 2005
    Posts
    15,598
    Thanks
    1421
    Quote Originally Posted by chr91 View Post
    Thanks so much for your time and effort to help. I understand it now.

    I have a similar question though. It's the same question apart from if f(x) tends to L as x tends to a, then |f(x)| tends to |L| as x tends to a..


    Is the format exactly the same but instead of epsilon / 2, we have mod of epsilon?
    epsilon is defined as positive so that would really make no sense.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Proving Limits
    Posted in the Calculus Forum
    Replies: 11
    Last Post: July 30th 2011, 09:09 AM
  2. Proving limits
    Posted in the Differential Geometry Forum
    Replies: 10
    Last Post: March 5th 2010, 04:07 AM
  3. Proving Limits
    Posted in the Calculus Forum
    Replies: 7
    Last Post: March 5th 2010, 02:38 AM
  4. [SOLVED] proving limits
    Posted in the Calculus Forum
    Replies: 2
    Last Post: October 4th 2009, 12:46 PM
  5. Proving limits help please???
    Posted in the Calculus Forum
    Replies: 3
    Last Post: January 29th 2009, 10:10 AM

Search Tags


/mathhelpforum @mathhelpforum