Results 1 to 4 of 4

Math Help - Confusion with limits

  1. #1
    Member rowe's Avatar
    Joined
    Jul 2009
    Posts
    89

    Confusion with limits

    \lim_{x \to -3} \frac{t^2-9}{2t^2 +7t+3} will equal 0 if you apply the usual limit laws and consider the limit as t tends to -3 in the numerator.

    Factorisation of the numerator and denominator gives us \frac{t-3}{2t+1}, which gives us the correct limit \frac{6}{5}. However, I only know this because I looked at the graph.

    It seems arbitrary that a simplification should produce a different result, if the expression is equal. How can I possibly tell whether a function has a limit at a particular point unless I do the "correct" manipulations? There must be some strict method I can apply to every function to ensure I know I am taking limits correctly.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member
    Joined
    Jun 2009
    From
    United States
    Posts
    676
    Thanks
    19
    Quote Originally Posted by rowe View Post
    \lim_{x \to -3} \frac{t^2-9}{2t^2 +7t+3} will equal 0 if you apply the usual limit laws and consider the limit as t tends to -3 in the numerator.

    Factorisation of the numerator and denominator gives us \frac{t-3}{2t+1}, which gives us the correct limit \frac{6}{5}. However, I only know this because I looked at the graph.

    It seems arbitrary that a simplification should produce a different result, if the expression is equal. How can I possibly tell whether a function has a limit at a particular point unless I do the "correct" manipulations? There must be some strict method I can apply to every function to ensure I know I am taking limits correctly.
    The function \frac{t^2-9}{2t^2 +7t+3} is not defined at x=-3 because the denominator is zero. You can't have a zero denominator. When you set x=-3 you should have obtained \frac{0}{0} which is undefined. Whenever this happens, you know you need to perform some manipulations since the direct substitution property of limits only applies when the x-value is in the domain of the function. In general

    \lim_{x->a}f(x)=f(a) if a is in D_f
    Last edited by adkinsjr; January 7th 2010 at 01:47 PM.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member rowe's Avatar
    Joined
    Jul 2009
    Posts
    89
    I gotcha, thanks.

    So what about this one:

    \lim_{t \to 0} \frac{1}{t \sqrt{(1+t)}}-\frac{1}{t}

    I've simplified it to

    \lim_{t \to 0} \frac{1-\sqrt{(1+t)}}{t\sqrt{1+t}},

    But tricks like rationalizing the numerator doesn't seem to work. Any ideas?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    11,814
    Thanks
    703
    Hello, =rowe!

    So what about this one: . \lim_{t \to 0} \frac{1}{t \sqrt{1+t}}-\frac{1}{t}

    I've simplified it to: . \lim_{t \to 0} \frac{1-\sqrt{1+t}}{t\sqrt{1+t}},

    But tricks like rationalizing the numerator doesn't seem to work. . Yes, it does . . .
    Multiply by \frac{1+\sqrt{1+t}}{1+\sqrt{1+t}}

    . . \frac{1-\sqrt{1+t}}{t\sqrt{1+t}}\cdot\frac{1+\sqrt{1+t}}{1  +\sqrt{1+t}} \;=\;\frac{1 - (1+t)}{t\sqrt{1+t}\,(1+\sqrt{t+1})} . =\;\frac{-t}{t\sqrt{1+t}\,(1+\sqrt{1+t})} \;=\;\frac{-1}{\sqrt{1+t}\,(1 + \sqrt{1+t})}


    Then: . \lim_{t\to0}\frac{-1}{\sqrt{1+t}\,(1 + \sqrt{1+t})} \;=\;\frac{-1}{\sqrt{1+0}\,(1 + \sqrt{1+0})} \;=\;\frac{-1}{(1)(2)} \;=\;-\frac{1}{2}<br /> <br /> <br />


    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. confusion
    Posted in the Statistics Forum
    Replies: 2
    Last Post: August 15th 2009, 02:17 AM
  2. confusion?
    Posted in the Math Topics Forum
    Replies: 1
    Last Post: August 8th 2009, 10:03 AM
  3. A bit of confusion w/ limits
    Posted in the Calculus Forum
    Replies: 1
    Last Post: December 28th 2008, 11:31 AM
  4. a confusion
    Posted in the Advanced Statistics Forum
    Replies: 1
    Last Post: November 24th 2008, 01:59 AM
  5. Confusion
    Posted in the Geometry Forum
    Replies: 2
    Last Post: October 21st 2006, 03:47 AM

Search Tags


/mathhelpforum @mathhelpforum