Results 1 to 3 of 3

Math Help - Would like Help with Calc problem!

  1. #1
    Newbie
    Joined
    Aug 2010
    Posts
    1

    Would like Help with Calc problem!

    Here's the problem
    limit (sqrt(x^2+5x)-x)
    x->[infinity]

    The answer is (according to two websites and my calculator) 5/2.

    Here's the work I've done so far.

    [Multiply the conjugate]
    (sqrt(x^2+5x)-x)/1 * (sqrt(x^2+5x)+x)/(sqrt(x^2+5x)+x)

    x^2+5x-x^2/(sqrt(x^2+5x)+x) = 5x/(sqrt(x^2+5x)+x)

    and I don't know how to get farther than that. Please help!
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member
    Joined
    Jan 2008
    Posts
    588
    Thanks
    87
    Quote Originally Posted by MisanthropicMan View Post
    Here's the problem
    limit (sqrt(x^2+5x)-x)
    x->[infinity]

    The answer is (according to two websites and my calculator) 5/2.

    Here's the work I've done so far.

    [Multiply the conjugate]
    (sqrt(x^2+5x)-x)/1 * (sqrt(x^2+5x)+x)/(sqrt(x^2+5x)+x)

    x^2+5x-x^2/(sqrt(x^2+5x)+x) = 5x/(sqrt(x^2+5x)+x)

    and I don't know how to get farther than that. Please help!
    Continuing from where you left off:
     <br />
	\displaystyle\lim_{x\to\infty}   \frac{5x}{\sqrt{x^2+5x}+x}

    Here is a nice little trick: factor out an x from the denominator:

    =  \displaystyle\lim_{x\to\infty}<br />
\frac{5x}{x\left(\frac{\sqrt{x^2+5x}}{x} + 1\right)}

    \displaystyle\lim_{x\to\infty} <br />
= \frac{5}{\frac{\sqrt{x^2+5x}}{x} + 1}

     =\displaystyle\lim_{x\to\infty} <br />
\frac{5}{\sqrt{\frac{x^2+5x}{x^2}} + 1}

    As everything else is constant:

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

    I trust you can continue from here?
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor

    Joined
    Apr 2005
    Posts
    15,707
    Thanks
    1470
    Quote Originally Posted by Gusbob View Post
    Continuing from where you left off:
     <br />
	\displaystyle\lim_{x\to\infty}   \frac{5x}{\sqrt{x^2+5x}+x}

    Here is a nice little trick: factor out an x from the denominator:

    =  \displaystyle\lim_{x\to\infty}<br />
\frac{5x}{x\left(\frac{\sqrt{x^2+5x}}{x} + 1\right)}
    Or, same thing, divide both numerator and denominator by x.

    \displaystyle\lim_{x\to\infty} <br />
= \frac{5}{\frac{\sqrt{x^2+5x}}{x} + 1}

     =\displaystyle\lim_{x\to\infty} <br />
\frac{5}{\sqrt{\frac{x^2+5x}{x^2}} + 1}

    As everything else is constant:

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

    I trust you can continue from here?
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Max Min Calc Problem
    Posted in the Calculus Forum
    Replies: 4
    Last Post: September 27th 2011, 04:28 AM
  2. Calc. Problem
    Posted in the Calculus Forum
    Replies: 1
    Last Post: December 11th 2009, 02:11 PM
  3. calc 3 problem...
    Posted in the Calculus Forum
    Replies: 6
    Last Post: June 4th 2008, 02:28 PM
  4. Help with Calc problem....
    Posted in the Calculus Forum
    Replies: 4
    Last Post: May 15th 2008, 01:36 PM
  5. help-calc problem
    Posted in the Calculus Forum
    Replies: 4
    Last Post: April 18th 2006, 11:46 PM

Search Tags


/mathhelpforum @mathhelpforum