Results 1 to 3 of 3

Math Help - limits

  1. #1
    Junior Member
    Joined
    Jan 2010
    Posts
    47

    limits

    lim as x approaches 0  \frac{1}{x^4} - \frac{1}{x^2}

    I know im suppose to use l'hopitals rule but i dont see how to use it.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member Aryth's Avatar
    Joined
    Feb 2007
    From
    USA
    Posts
    651
    Thanks
    1
    Awards
    1
    It's actually easier than you think, let's look at your problem:

    \lim_{x \to 0} \left(\frac{1}{x^4} - \frac{1}{x^2}\right)

    Recall the distribution over subtraction property to get:

    \lim_{x \to 0}\frac{1}{x^4} - \lim_{x \to 0} \frac{1}{x^2}

    Now, you can solve these one at a time with L'Hopital and then subtract them, I'll work out the second one:

    \lim_{x \to 0}\frac{1}{x^2}

    Apply L'Hopital's rule:

    \lim_{x \to 0} \frac{0}{2x} = 0

    I think you can do the other one.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Super Member General's Avatar
    Joined
    Jan 2010
    From
    Kuwait
    Posts
    562
    Quote Originally Posted by Aryth View Post
    It's actually easier than you think, let's look at your problem:

    \lim_{x \to 0} \left(\frac{1}{x^4} - \frac{1}{x^2}\right)

    Recall the distribution over subtraction property to get:

    \lim_{x \to 0}\frac{1}{x^4} - \lim_{x \to 0} \frac{1}{x^2}

    Now, you can solve these one at a time with L'Hopital and then subtract them, I'll work out the second one:

    \lim_{x \to 0}\frac{1}{x^2}

    Apply L'Hopital's rule:

    \lim_{x \to 0} \frac{0}{2x} = 0

    I think you can do the other one.
    You can not do that.
    \lim_{x\to 0} \frac{1}{x^2} is not one of the L`Hospital`s Rule cases.
    Also, if \lim_{x\to a} f(x) and \lim_{x\to a} g(x) are infinite, then (in general):
    \lim_{x\to a} \left( f(x) - g(x) \right) {\color{red} \neq } \lim_{x\to a} f(x) - \lim_{x\to a} g(x)

    For OP's problem:
    \lim_{x\to 0} \left( \frac{1}{x^4} - \frac{1}{x^2} \right) = \lim_{x\to 0} \frac{1}{x^2} \left( \frac{1}{x^2} - 1 \right) = \infty ( \infty - 1 ) = \infty
    and this means the y-axis is a vertical asymptote for the graph of the function f(x)=\frac{1}{x^4} - \frac{1}{x^2}.
    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, 05:34 PM
  2. Function limits and sequence limits
    Posted in the Differential Geometry Forum
    Replies: 0
    Last Post: April 26th 2009, 01:45 PM
  3. HELP on LIMITS
    Posted in the Calculus Forum
    Replies: 4
    Last Post: September 23rd 2008, 11:17 PM
  4. Limits
    Posted in the Calculus Forum
    Replies: 5
    Last Post: September 21st 2008, 10:52 PM
  5. [SOLVED] [SOLVED] Limits. LIMITS!
    Posted in the Calculus Forum
    Replies: 1
    Last Post: February 25th 2008, 10:41 PM

Search Tags


/mathhelpforum @mathhelpforum