Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

How to Find This Limit?

  1. Jan 17, 2016 #1
    How to find the

    ##
    \lim_{x \to \infty} (\sqrt{ax^2 + bx + c} - \sqrt{ax^2 + px + q}) = \frac{b - p}{2 \sqrt{a}}
    ##

    ???

    Here is what I get

    ##
    \lim_{x \to \infty} (\sqrt{ax^2 + bx + c} - \sqrt{ax^2 + px + q})
    ##
    ##
    = \sqrt{\lim_{x \to \infty} ax^2 + \lim_{x \to \infty} bx + \lim_{x \to \infty} c} - \sqrt{\lim_{x \to \infty} ax^2 + \lim_{x \to \infty} px + \lim_{x \to \infty} q}
    ##
    ##
    =\sqrt{\infty + \infty + c} - \sqrt{\infty + \infty + q}
    ##
    ##
    = \sqrt{\infty} - \sqrt{\infty}
    ##
    ##
    = \infty
    ##
     
  2. jcsd
  3. Jan 17, 2016 #2

    Samy_A

    User Avatar
    Science Advisor
    Homework Helper

    ∞-∞ is undefined, so the last step is wrong.

    Hint: ##(\sqrt y- \sqrt z)(\sqrt y+ \sqrt z)=y-z##
     
  4. Jan 17, 2016 #3

    WWGD

    User Avatar
    Science Advisor
    Gold Member

    OP: Or you can complete the square inside each of the radicals and ignore the constant terms.
     
  5. Jan 17, 2016 #4
    Do you mean

    ##(\sqrt{ax^2 + bx + c} - \sqrt{ax^2 + px + q})(\sqrt{ax^2 + bx + c} + \sqrt{ax^2 + px + q})##
    ##= (ax^2 + bx + c)-(ax^2 + px + q)##

    ???????
     
  6. Jan 17, 2016 #5

    Samy_A

    User Avatar
    Science Advisor
    Homework Helper

    Yes, that is correct (although not exactly what I meant).
    You can further simplify the result.
    After that, use this other mathematical identity: ##A=\frac{A*B}{B}##, where ##B\neq 0##.
     
    Last edited: Jan 17, 2016
  7. Jan 17, 2016 #6

    WWGD

    User Avatar
    Science Advisor
    Gold Member

    No, he surely meant (sorry to speak for you Samy_A) that you must both multiply and divide by this expression , otherwise , if you only multiply and don't divide, you are changing the original expression for which you want to find the limit.
     
  8. Jan 17, 2016 #7
    CAn we just go to the point how to find the result? Please tell me step-by-step, I am confuse.
     
  9. Jan 17, 2016 #8

    Mark44

    Staff: Mentor

    That's not how it works here. We will steer you in the right direction, but you need to do the majority of the work.
    Instead of multiplying by ##(\sqrt{ax^2 + bx + c} + \sqrt{ax^2 + px + q})##, multiply by 1, in the form of ##(\sqrt{ax^2 + bx + c} + \sqrt{ax^2 + px + q})## over itself. Multiplying by 1 doesn't change the value of an expression, so is always valid to do.
     
  10. Jan 17, 2016 #9
    Isn't multplying by 1 (in the form of ##(\sqrt{ax^2 + bx + c} + \sqrt{ax^2 + px + q})##) just make more complex because the denominator is in the form of square root?

    ##(\sqrt{ax^2 + bx + c} - \sqrt{ax^2 + pc + q}) × \frac{\sqrt{ax^2 + bx + c} + \sqrt{ax^2 + px + q}}{\sqrt{ax^2 + bx + c} + \sqrt{ax^2 + px + q}}##

    ##= \frac{(ax^2 + bx + c) - (ax^2 + px + q)}{\sqrt{ax^2 + bx + c} + \sqrt{ax^2 + px + q}}##
    ##=\frac{bx + c - px - q}{\sqrt{ax^2 + bx + c} + \sqrt{ax^2 + px + q}}##
    ##=\frac{bx - px + c - q}{\sqrt{ax^2 + bx + c} + \sqrt{ax^2 + px + q}}##
    ##=\frac{(b - p)x + c - q}{\sqrt{ax^2 + bx + c} + \sqrt{ax^2 + px + q}}##
     
  11. Jan 17, 2016 #10

    Samy_A

    User Avatar
    Science Advisor
    Homework Helper

    Ok, now look at what you have.
    Let's first make an informal argument. As you need the limit for ##x \to + \infty##, we can consider only the highest order terms in both numerator and denominator (because for large ##x## they are much larger than the other terms).
    That gives ##\frac{(b-p)x}{\sqrt{ax^2}+\sqrt{ax^2}}=\frac{(b-p)x}{2x\sqrt{a}}=\frac{b-p}{2\sqrt{a}}##.

    Now all you have to do is make this informal argument formally correct.
     
  12. Jan 17, 2016 #11
    How do you get ##\frac{(b-p)x}{\sqrt{ax^2}+\sqrt{ax^2}}##? What happen to the c - q in the numerator and bx + c, and px + q in the denominator?
     
  13. Jan 17, 2016 #12

    Samy_A

    User Avatar
    Science Advisor
    Homework Helper

    It's explained in the previous post.

    Take the numerator: ##(b-p)x+(c-q)##. When ##x## gets very large, the value of the numerator will mainly be determined by ##(b-p)x##. The ##c-q## term is constant, and will become negligible compared to ##(b-p)x## for large ##x##. Similarly for the denominator.
    That's the informal computation I did in the previous post, and you can see that the result is the searched limit.

    Now convert this informal argument into a formal one. Hint: Divide both numerator and denominator by ##x##, take the ##1/x## in the denominator into the square roots, and then take the limit.
     
  14. Jan 17, 2016 #13
    I still don't understand.

    Btw, this is the calculus textbook I use to learn about limit:

    << Illegal download link deleted by Moderator >>

    Can someone please tell me what page of above book about limit x approach infinity similar to above problem?
     
    Last edited by a moderator: Jan 17, 2016
  15. Jan 17, 2016 #14

    Samy_A

    User Avatar
    Science Advisor
    Homework Helper

    Divide both numerator and denominator by ##x##:
    $$ \frac{(b - p)x + c - q}{\sqrt{ax^2 + bx + c} + \sqrt{ax^2 + px + q}}=\frac{(b - p) + \frac{c - q}{x}}{\frac{1}{x}\sqrt{ax^2 + bx + c} + \frac{1}{x}\sqrt{ax^2 + px + q}}$$
    Now take the ##1/x## in the denominator into the square roots, and then take the limit.
     
  16. Jan 17, 2016 #15

    Erland

    User Avatar
    Science Advisor

    askor, the most similar example in your book is Example 9. p. 107, but it is a lot simpler than this one. Exercises 80-86, p. 116 are af the same type as your example. If you have a solution manual you might look up the solutions of these.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: How to Find This Limit?
  1. Finding the Limit (Replies: 1)

  2. Finding Limit (Replies: 11)

  3. Finding a limit (Replies: 6)

Loading...