Page 2 of 3 FirstFirst 123 LastLast
Results 16 to 30 of 33

Math Help - Limit of A Sequence

  1. #16
    Super Member redsoxfan325's Avatar
    Joined
    Feb 2009
    From
    Swampscott, MA
    Posts
    943
    Quote Originally Posted by RockHard View Post
    Can anyone provide me with another example before I deem this thread solved
    Prove that \lim_{n\to\infty}\left(\sqrt{n^2+1}-n\right)=0
    Follow Math Help Forum on Facebook and Google+

  2. #17
    Super Member redsoxfan325's Avatar
    Joined
    Feb 2009
    From
    Swampscott, MA
    Posts
    943
    Here's a trickier one.

    Prove \lim_{n\to\infty}n^2e^{-n}=0

    Hint: Rewrite n^2e^{-n} as e^{\ln(n^2)}e^{-n}=e^{2\ln(n)-n} and then try to put an upper bound on 2\ln(n) (that's easy to work with).

    Note that if you follow the hint, your value for N will look something like N=\max\{c, f(\epsilon)\} for some number c and some expression depending on \epsilon.

    I hope what I wrote wasn't too confusing.
    Follow Math Help Forum on Facebook and Google+

  3. #18
    Member
    Joined
    Nov 2009
    Posts
    186
    Whew, thanks man! Ill write these down and get to it, those look like a solid a example to work with, I hope I can get them correct or close to it.
    Follow Math Help Forum on Facebook and Google+

  4. #19
    Super Member redsoxfan325's Avatar
    Joined
    Feb 2009
    From
    Swampscott, MA
    Posts
    943
    The second one is (I think) significantly more difficult than the first, so don't worry if you have a hard time.
    Follow Math Help Forum on Facebook and Google+

  5. #20
    Member
    Joined
    Nov 2009
    Posts
    186
    Not gonna lie I am a tad confused with the first one because of the extra term in the equation I know I am probalby over complicating it
    Follow Math Help Forum on Facebook and Google+

  6. #21
    Super Member redsoxfan325's Avatar
    Joined
    Feb 2009
    From
    Swampscott, MA
    Posts
    943
    Quote Originally Posted by RockHard View Post
    Not gonna lie I am a tad confused with the first one because of the extra term in the equation I know I am probalby over complicating it
    Try multiplying by \frac{\sqrt{n^2+1}+n}{\sqrt{n^2+1}+n} to end up with \frac{1}{\sqrt{n^2+1}+n}.
    Follow Math Help Forum on Facebook and Google+

  7. #22
    Member
    Joined
    Nov 2009
    Posts
    186
    I know this is a dumb question but I have to be sure before I attempt the problem, is everything under the square root or is it just n^2 + 1 also isnt it - n or I missed something in your hint
    Follow Math Help Forum on Facebook and Google+

  8. #23
    Super Member redsoxfan325's Avatar
    Joined
    Feb 2009
    From
    Swampscott, MA
    Posts
    943
    Quote Originally Posted by RockHard View Post
    I know this is a dumb question but I have to be sure before I attempt the problem, is everything under the square root or is it just n^2 + 1 also isnt it - n or I missed something in your hint
    Just n^2+1.

    \sqrt{n^2+1}-n\cdot\frac{\sqrt{n^2+1}+n}{\sqrt{n^2+1}+n}=\frac{  n^2+1-n^2}{\sqrt{n^2+1}+n}=\frac{1}{\sqrt{n^2+1}+n}
    Follow Math Help Forum on Facebook and Google+

  9. #24
    Member
    Joined
    Nov 2009
    Posts
    186
    Man, my algebra never has been the greatest and I always get stuck on some problems that should be relatively easy because I second guess what I do if it is the correct way or not also, I know this horrible but what was the algebraic reason for multiplying by what you did? to make it fit the condition of absolute value a bit easier? but ill show a bit of work to where I am at

    \frac{1}{\sqrt{n^2+1}+n} < \epsilon

    1 < {\epsilon}{{\sqrt{n^2+1}+n}}

    Usually around steps like these I cannot determine If I am doing it correctly and make silly mistakes

    Could I even do

     {\sqrt{n^2+1}+n} > \frac{1}{\epsilon}

    Edit: I must continue so my erroneous ways can be corrected, even if it hurts, not literally.

     {n^2+1+n^2} > ({\frac{1}{\epsilon}})^2

     {2n^2+1} > ({\frac{1}{\epsilon}})^2

     {2n^2} > ({\frac{1}{\epsilon}})^2 - 1

    Stuck
    Last edited by RockHard; November 17th 2009 at 07:05 PM.
    Follow Math Help Forum on Facebook and Google+

  10. #25
    MHF Contributor Drexel28's Avatar
    Joined
    Nov 2009
    From
    Berkeley, California
    Posts
    4,563
    Thanks
    21
    Quote Originally Posted by RockHard View Post
    Man, my algebra never has been the greatest and I always get stuck on some problems that should be relatively easy because I second guess what I do if it is the correct way or not also, I know this horrible but what was the algebraic reason for multiplying by what you did? to make it fit the condition of absolute value a bit easier? but ill show a bit of work to where I am at

    \frac{1}{\sqrt{n^2+1}+n} < \epsilon

    1 < {\epsilon}{{\sqrt{n^2+1}+n}}

    Usually around steps like these I cannot determine If I am doing it correctly and make silly mistakes

    Could I even do

     {\sqrt{n^2+1}+n} > \frac{1}{\epsilon}
    Merely notice that \frac{1}{\sqrt{n^2+1}+n}\le\frac{1}{n}
    So letting n>\frac{1}{\varepsilon} does the trick.
    Follow Math Help Forum on Facebook and Google+

  11. #26
    Super Member redsoxfan325's Avatar
    Joined
    Feb 2009
    From
    Swampscott, MA
    Posts
    943
    Quote Originally Posted by RockHard View Post
    Man, my algebra never has been the greatest and I always get stuck on some problems that should be relatively easy because I second guess what I do if it is the correct way or not also, I know this horrible but what was the algebraic reason for multiplying by what you did? to make it fit the condition of absolute value a bit easier? but ill show a bit of work to where I am at

    \frac{1}{\sqrt{n^2+1}+n} < \epsilon

    1 < {\epsilon}{{\sqrt{n^2+1}+n}}

    Usually around steps like these I cannot determine If I am doing it correctly and make silly mistakes

    Could I even do

     {\sqrt{n^2+1}+n} > \frac{1}{\epsilon}

    Edit: I must continue so my erroneous ways can be corrected, even if it hurts, not literally.

     {n^2+1+n^2} > ({\frac{1}{\epsilon}})^2

     {2n^2+1} > ({\frac{1}{\epsilon}})^2

     {2n^2} > ({\frac{1}{\epsilon}})^2 - 1

    Stuck
    What Drexel did was what I had in mind for the solution. If you do it this, way, you're going to get something like:

    \sqrt{n^2+1}+n>\frac{1}{\epsilon}\implies 2n^2+1{\color{red}+2n\sqrt{n^2+1}}>(1/\epsilon)^2\implies 2n\sqrt{n^2+1}>(1/\epsilon)^2-1-2n^2 \implies 4n^2(n^2+1)>[(1/\epsilon)^2-1-2n^2]^2

    and solving that equation for n will give you a nasty quadratic.
    Follow Math Help Forum on Facebook and Google+

  12. #27
    Member
    Joined
    Nov 2009
    Posts
    186
    Hmm, trying to make sense of what Drexel did, it is like it magically disappeared that nasty square root.
    Follow Math Help Forum on Facebook and Google+

  13. #28
    MHF Contributor Drexel28's Avatar
    Joined
    Nov 2009
    From
    Berkeley, California
    Posts
    4,563
    Thanks
    21
    Quote Originally Posted by RockHard View Post
    Hmm, trying to make sense of what Drexel did, it is like it magically disappeared that nasty square root.
    All you have to realize is that saying \frac{1}{\text{whatever}}<\frac{1}{\text{whatever else}} assuming positivity is that \text{whatever}>\text{whatever else}. Now just note that n^2+1>n^2\implies \sqrt{n^2+1}>\sqrt{n^2}=n\implies \sqrt{n^2+1}+n>2n\implies \sqrt{n^2+1}+n>n
    Follow Math Help Forum on Facebook and Google+

  14. #29
    Member
    Joined
    Nov 2009
    Posts
    186
    So you can you always do this sort of thing, if it easy to find? What theorem is this called? There so many theorems and idea's you dont know when to apply what
    Follow Math Help Forum on Facebook and Google+

  15. #30
    MHF Contributor Drexel28's Avatar
    Joined
    Nov 2009
    From
    Berkeley, California
    Posts
    4,563
    Thanks
    21
    Quote Originally Posted by RockHard View Post
    So you can you always do this sort of thing, if it easy to find? What theorem is this called? There so many theorems and idea's you dont know when to apply what
    Theorem? This is just a property of the real numbers since they are an ordered field? There isn't a theorem here, it's just the way it is.
    Follow Math Help Forum on Facebook and Google+

Page 2 of 3 FirstFirst 123 LastLast

Similar Math Help Forum Discussions

  1. limit of sequence
    Posted in the Calculus Forum
    Replies: 3
    Last Post: June 28th 2011, 05:09 PM
  2. Replies: 2
    Last Post: October 26th 2010, 10:23 AM
  3. limit of a sequence
    Posted in the Calculus Forum
    Replies: 2
    Last Post: March 24th 2010, 05:08 PM
  4. Limit of a Sequence
    Posted in the Differential Geometry Forum
    Replies: 30
    Last Post: March 6th 2010, 09:10 AM
  5. Limit of sequence
    Posted in the Calculus Forum
    Replies: 4
    Last Post: March 12th 2008, 03:16 PM

Search Tags


/mathhelpforum @mathhelpforum