1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Finding the limit as x-> infinity: [sqrt(5 + 5x^2)]/(5 + 7x)

  1. Jan 10, 2013 #1
    Finding the limit as x--> infinity: [sqrt(5 + 5x^2)]/(5 + 7x)

    1. lim[√(5 + 5x^2)]/(5 +7x)
    x→∞

    2. Alright, I thought I would have first find the largest exponent of x in the denominator.

    In this case, the largest exponent is x^1.
    The next step is to divide every term by x^1.

    Since I cannot divide something in a square root by x, I thought I COULD multiply it by
    √(1/x^2). That's the same thing as dividing by x.

    So, this is what I have:

    [√(1/x^2)*√(5 + 5x^2)]/(5/x + 7x/x)

    = [ √(5/(x^2) + 5x^2/x^2) / ((5/x) + 7) ] * (1/x^2)/(1/x)

    = √((5/x^2) + 5) / ((5/x) + 7)

    Then I thought if you substitute infinity for x here, then the (5/x^2) and the 5/x both equal 0.

    So, it's √((5/0) + 5) / (0 + 7)

    = √(5)/7

    .....This is not the right answer... =_=
    Could you find my mistakes?
    Thank you so much!
     
  2. jcsd
  3. Jan 10, 2013 #2

    symbolipoint

    User Avatar
    Homework Helper
    Education Advisor
    Gold Member

    Re: Finding the limit as x--> infinity: [sqrt(5 + 5x^2)]/(5 + 7x)

    I was able to transform the expression to:
    EDIT: The TEX was just too tedious correct after making it wrong, so I am corrected my mistake in plain text:
    (sqrt((5/x^2)+5))/((5/x)+7)

    Upon letting x increase without end, I found the same limit as you did,
    [itex]\frac{\sqrt{5}}{7}[/itex]
     
  4. Jan 10, 2013 #3
    Re: Finding the limit as x--> infinity: [sqrt(5 + 5x^2)]/(5 + 7x)

    You're answer is correct:

    [tex]\lim_{x\rightarrow+\infty}\frac{\sqrt{5x^{2}+5}}{7x+5}=\lim_{x\rightarrow+\infty}\sqrt{\frac{5x^{2}+5}{(7x+5)^{2}}}=\lim_{x \rightarrow +\infty} \sqrt{\frac{5x^2+5}{49x^2+70x+25}}=\lim_{x \to +\infty}\sqrt{\frac{5x^2}{49x^2}}=\frac{\sqrt{5}}{7}[/tex]
     
    Last edited by a moderator: Jan 10, 2013
  5. Jan 10, 2013 #4

    symbolipoint

    User Avatar
    Homework Helper
    Education Advisor
    Gold Member

    Re: Finding the limit as x--> infinity: [sqrt(5 + 5x^2)]/(5 + 7x)

    Worked solution, because it may be somewhat different from LoLeeTa and mtayab1994.
    (Attached)
     

    Attached Files:

  6. Jan 10, 2013 #5
    Re: Finding the limit as x--> infinity: [sqrt(5 + 5x^2)]/(5 + 7x)

    Thanks so much symbolipoint and mtayab1994!

    You guys are awesome! :)

    There were 2 aspects to the computer hw problem.
    One part was as x→∞, and the other was as x→-∞

    So it actually counted the √(5)/7 right for x→∞

    BUT for x→-∞, the answer had to be -√(5)/7!

    Thanks! :D
     
  7. Jan 10, 2013 #6

    Mark44

    Staff: Mentor

    Re: Finding the limit as x--> infinity: [sqrt(5 + 5x^2)]/(5 + 7x)

    That's because ##\sqrt{x^2}## = -x if x < 0. Keep in mind that the square root of a nonnegative number is nonnegative, by definition.

    In essence you are factoring x2 out of the terms insided the radical, and bringing it out of the radical as -x.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Finding the limit as x-> infinity: [sqrt(5 + 5x^2)]/(5 + 7x)
Loading...