Results 1 to 3 of 3

Math Help - Ugh...Algebra is harder than Calculus

  1. #1
    Member
    Joined
    Dec 2005
    Posts
    117

    Ugh...Algebra is harder than Calculus

    Need help with some algebraic simplification. This is an example from my calculus book asking to find the vertical and horizontal asymptotes. They divided the numerator and denominator by x. I don't understand how they they got to the second step after dividing by x and simplifying. Can someone explain this? Radicals screw everything up for me.

    \lim_{x\to\infty}\frac{\sqrt{2x^2+1}}{3x-5} =
    \lim_{x\to\infty}\frac{\sqrt{2+(1/x^2)}}{3-(5/x)}
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Member
    Joined
    Jun 2006
    From
    San Diego
    Posts
    101
    I'll do what I can.

    First, you know things like that \sqrt{x}\cdot\sqrt{y}=\sqrt{xy}, right? I hope so because I will use little rules like that in this demonstration.

    Start with what you had:

    \lim_{x\to\infty} \frac{\sqrt{2x^2+1}}{3x-5}.

    Dividing the top and bottom by x is like multiplying the top and bottom by \frac{1}{x}.

    \lim_{x\to\infty} \frac{\left(\frac{1}{x}\right)\sqrt{2x^2+1}}{\left  (\frac{1}{x}\right)(3x-5)}

    Well \frac{1}{x} = \sqrt{\frac{1}{x^2}}. This isn't normally completely correct because the right side is always positive while the left could be positive or negative, but we are allowed in this case since we are taking the limit to positive infinity. Therefore at some point we will only care about positive values of x (and extremely high ones, at that) and can ignore cases where this equality isn't necessarily true.

    Make ths substitution to put this back into what we're working with:

    \lim_{x\to\infty} \frac{\sqrt{\frac{1}{x^2}} \cdot \sqrt{2x^2+1}}{\left(\frac{1}{x}\right)(3x-5)}

    Now apply the little rule I mentioned at the top (and also distribute the bottom).

    \lim_{x\to\infty} \frac{\sqrt{\left(\frac{1}{x^2}\right)(2x^2+1)}}{3-\frac{5}{x}}

    Distribute inside the radical to get that which was desired:

    \lim_{x\to\infty} \frac{\sqrt{2+(1/x^2)}}{3-(5/x)}.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member
    Joined
    Dec 2005
    Posts
    117
    thank you!
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Harder Integration
    Posted in the Calculus Forum
    Replies: 4
    Last Post: May 9th 2009, 01:26 PM
  2. harder gradient
    Posted in the Geometry Forum
    Replies: 1
    Last Post: June 18th 2008, 03:49 AM
  3. a little harder
    Posted in the Statistics Forum
    Replies: 1
    Last Post: October 29th 2007, 11:02 PM
  4. Here's another (harder) one
    Posted in the Pre-Calculus Forum
    Replies: 2
    Last Post: August 14th 2007, 10:50 PM
  5. I think I am making this harder
    Posted in the Calculus Forum
    Replies: 2
    Last Post: June 4th 2006, 09:28 AM

Search Tags


/mathhelpforum @mathhelpforum