Results 1 to 8 of 8

Math Help - problems on limits of sequences

  1. #1
    Junior Member
    Joined
    Jul 2006
    Posts
    73

    problems on limits of sequences

    1)
    Determine whether or not the following limits exist. Justify your answers:
    A) Lim x->0+ (1/x)
    B) Lim x->0+ |sin(1/x)|
    C) Lim x->0+ xsin(1/x)

    2)
    Determine the following limits:
    A) Lim x->1 (x^3+5)/(x^2+2)
    B) Lim x->1 (sqrt(x)-1)/(x-1)
    C)Lim x->0 (x^2 + 4x)/(x^2+2x)
    D) Lim x->0 (sqrt(4+x) - 2)/x
    E) lim x->0- 4x/|x|
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member
    Joined
    Oct 2006
    Posts
    679
    Awards
    1
    Quote Originally Posted by luckyc1423 View Post
    1)
    Determine whether or not the following limits exist. Justify your answers:
    A) Lim x->0+ (1/x)
    B) Lim x->0+ |sin(1/x)|
    C) Lim x->0+ xsin(1/x)

    2)
    Determine the following limits:
    A) Lim x->1 (x^3+5)/(x^2+2)
    B) Lim x->1 (sqrt(x)-1)/(x-1)
    C)Lim x->0 (x^2 + 4x)/(x^2+2x)
    D) Lim x->0 (sqrt(4+x) - 2)/x
    E) lim x->0- 4x/|x|
    I am having somewhat of a difficult time understanding your notation for the first section of problems. What are you trying to indicate by using the plus (+) sign? Also for problems D which part is under the square root? sqrt(4+x)-2? Why not write 2-x? Some clarification would be sweet.

    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    Jul 2006
    Posts
    73
    The 0+ I think means you are approaching 0 from the positive direction

    the 4+x is the only thing in the sqrt

    sqrt(4+x)
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Junior Member
    Joined
    Jul 2006
    Posts
    73
    so its sqrt(4+x) then minus 2 and all of that divided by x
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Super Member
    Joined
    Oct 2006
    Posts
    679
    Awards
    1

    Re:

    Here is a Graph for #2 E. This is one of those indeterminate problems there for you must take the limit from both the right and left hand side. It is easy to see by looking the graph. The left-hand limit is -4, and just as you can tell the right hand limit is 4. Why? Because the function is zeroing in on the Y axis to -4 and 4 respectfully.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Super Member
    Joined
    Oct 2006
    Posts
    679
    Awards
    1

    Re:

    Quote Originally Posted by luckyc1423 View Post
    so its sqrt(4+x) then minus 2 and all of that divided by x
    Ohh...In this case it now big deal you just have to multiply by a factor of one. Any time you initial solution yields indeterminate 0/0 or inf/inf or -inf/-inf or -inf/inf or inf/-inf you need to factor or use algebra.

    Follow Math Help Forum on Facebook and Google+

  7. #7
    is up to his old tricks again! Jhevon's Avatar
    Joined
    Feb 2007
    From
    New York, USA
    Posts
    11,663
    Thanks
    3
    Quote Originally Posted by luckyc1423 View Post
    1)
    Determine whether or not the following limits exist. Justify your answers:
    A) Lim x->0+ (1/x)
    notice as we approach 1/x from the right, we have positive values. so our limit will be positive.

    now lim{x--> 0+} 1/x = +oo, since 1/x --> oo as x --> 0 and we have a positive limit


    B) Lim x->0+ |sin(1/x)|
    we saw above that as x-->0+, 1/x --> +oo

    so for lim{x-->0+} |sin(1/x)|, we have sin(1/x)-->sin(+oo) as x-->0+, now sine oscillates between 1 and -1 for it's maximum and minimum values. since we have |sin(1/x)| however, all negative values become positive. however, we still have oscillations as we change x. so the limit does not exist

    C) Lim x->0+ xsin(1/x)
    lim{x-->0+}xsin(1/x) = lim{x-->0+}x * lim{x-->0+}sin(1/x) = 0*lim{x-->0+}sin(1/x) = 0
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Super Member
    Joined
    Oct 2006
    Posts
    679
    Awards
    1

    Re:

    Ohh...Yea thanks Jhevon. Since Latex is down I couldn't tell why the plus (+) sign was being used. Duhh... Approach from the right. Hopefully they are still working on getting it fixed. Thanks!
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Sequences and limits
    Posted in the Differential Geometry Forum
    Replies: 3
    Last Post: October 1st 2011, 07:00 PM
  2. Limits and sequences
    Posted in the Calculus Forum
    Replies: 2
    Last Post: September 17th 2010, 01:51 AM
  3. limits of sequences
    Posted in the Calculus Forum
    Replies: 2
    Last Post: June 11th 2010, 12:16 AM
  4. Limits of sequences
    Posted in the Calculus Forum
    Replies: 10
    Last Post: May 12th 2010, 06:54 AM
  5. Sequences - Limits
    Posted in the Calculus Forum
    Replies: 9
    Last Post: May 25th 2008, 01:43 PM

Search Tags


/mathhelpforum @mathhelpforum