Results 1 to 6 of 6

Math Help - Limits Help

  1. #1
    tmd
    tmd is offline
    Newbie
    Joined
    Jul 2006
    Posts
    5

    Limits Help

    1. (2 + e^1/x) / (2 - e^1/x)
    What is the horizontal asymptotes?

    2. f(x) = { sin x/x, x cant be 0
    k, x = 0
    In order for f(x) to be continous at x=0, the value of k must be ?

    3. If a cant be 0, then Limit x->a (x^2 - a^2)/ (x^4 - a^4) is?

    Help please? thankyou
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Forum Admin topsquark's Avatar
    Joined
    Jan 2006
    From
    Wellsville, NY
    Posts
    10,212
    Thanks
    419
    Awards
    1
    Quote Originally Posted by tmd View Post
    (2 + e^1/x) / (2 - e^1/x)
    What is the horizontal asymptotes?
    I'm assuming this function is: (2 + e^{1/x})/(2 - e^{1/x})?
    Horizontal asymptotes occur when we take the function limit at x -> (+/-) infinity.

    So, as x -> - infinity, 1/x -> 0 so e^{1/x} -> 1 Thus your function approaches (2 + 1)/(2 - 1) = 2. Since this is a constant, we have a horizontal asymptote at y = 2 for x->-infinity.

    As x -> infinity we have the same behavior for 1/x so the function again has a horizontal asymptote at y = 2 for x->infinity.

    (Technically there are still two horizontal asymptotes, even though they are the same line.)

    -Dan
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Forum Admin topsquark's Avatar
    Joined
    Jan 2006
    From
    Wellsville, NY
    Posts
    10,212
    Thanks
    419
    Awards
    1
    Quote Originally Posted by tmd View Post
    f(x) = { sin x/x, x cant be 0
    k, x = 0
    In order for f(x) to be continous at x=0, the value of k must be ?
    Are you saying that for x not equal to zero f(x) = sin(x)/x and at x = 0 f(x) = k?

    For f(x) to be continous we must have that k = lim(x->0) sin(x)/x. There are various ways to find this limit. Do you know the power series expansion of sin(x)? If so, this is probably the simplest way to get this limit.

    For x close to 0
    sin(x) = x - (1/3!)x^3 + (1/5!)x^5 - ...

    So for x close to 0
    sin(x)/x is approximately [x - (1/3!)x^3 + (1/5!)x^5 - ...]/x = 1 - (1/3!)x^2 + (1/5!)x^4 - ...
    which is approximately 1. ( x^n << 1 for x close to 0 for all positive integers n)

    Thus k = 1.

    -Dan
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Forum Admin topsquark's Avatar
    Joined
    Jan 2006
    From
    Wellsville, NY
    Posts
    10,212
    Thanks
    419
    Awards
    1
    Quote Originally Posted by tmd View Post
    If a cant be 0, then Limit x->a (x^2 - a^2)/ (x^4 - a^4) is?
    x^4 - a^4 is the difference between two squares (x^2)^2 and (a^2)^2 so this factors as:
    x^4 - a^4 = (x^2 + a^2)(x^2 - a^2)

    so
    (x^2 - a^2)/(x^4 - a^4) = 1/(x^2 + a^2)

    So lim(x->a) (x^2 - a^2)/(x^4 - a^4) = 1/(a^2 + a^2) = 1/(2a^2)

    -Dan
    Last edited by topsquark; September 28th 2006 at 06:19 AM. Reason: I get it now! ;)
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Global Moderator

    Joined
    Nov 2005
    From
    New York City
    Posts
    10,616
    Thanks
    10
    Quote Originally Posted by topsquark View Post

    For x close to 0
    sin(x) = x - (1/3!)x^3 + (1/5!)x^5 - ...
    No need to do that.
    The limit of,
    sin(x)/x
    Is one of those students must know immediately.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Global Moderator

    Joined
    Nov 2005
    From
    New York City
    Posts
    10,616
    Thanks
    10
    Quote Originally Posted by tmd View Post
    1. (2 + e^1/x) / (2 - e^1/x)
    What is the horizontal asymptotes?
    Divide the numerator and demoninator by e^(1/x)
    Thus,
    (2e^(-1/x)+1)/(2e^(-1/x)-1)
    When x---> +oo
    The numerator is, 2(1)+1=3
    The denominator is, 2(1)-1=1 not equal to zero.
    Thus, by the quotient rule of limits this limit is, (3)/(1)=3
    That is one horizontal asymptote.
    ---
    Instead of dividing leave it the way it should be
    (2 + e^(1/x)) / (2 - e^(1/x))
    As x---> -oo
    The numerator is 2+1=3
    The denominator is 2-1=1 not equal to zero.
    Thus, by the qtuotient rule of limits this limit is (3)/(1)=3
    Also the same.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Using limits to find other limits
    Posted in the Calculus Forum
    Replies: 7
    Last Post: September 18th 2009, 06:34 PM
  2. Function limits and sequence limits
    Posted in the Differential Geometry Forum
    Replies: 0
    Last Post: April 26th 2009, 02:45 PM
  3. HELP on LIMITS
    Posted in the Calculus Forum
    Replies: 4
    Last Post: September 24th 2008, 12:17 AM
  4. Limits
    Posted in the Calculus Forum
    Replies: 5
    Last Post: September 21st 2008, 11:52 PM
  5. [SOLVED] [SOLVED] Limits. LIMITS!
    Posted in the Calculus Forum
    Replies: 1
    Last Post: February 25th 2008, 11:41 PM

Search Tags


/mathhelpforum @mathhelpforum