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Lim sq.rt x[sqrt x - sqrt (x-a)] x->infinity

  1. Jul 2, 2009 #1
    lim sq.rt x[sqrt x - sqrt (x-a)] x-->infinity

    I'm whacked outta solving these problems just hanging around for more than 2 hours for each question but can't solve it.I think it's out of my capability.LOL.
    I hope you will solve these problems.I need your help.The questions are...
    1) lim sq.rt x[sqrt x - sqrt (x-a)]
    x-->infinity

    2) lim [tan x - sin x]/x^3
    x--> 0

    Thanks in advance.
     
  2. jcsd
  3. Jul 2, 2009 #2

    dx

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    Re: limits

    You must show your attempt to receive help.
     
  4. Jul 2, 2009 #3

    HallsofIvy

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    Re: limits

    You say you've been working on it for two hours (which is not really that much) so show what you have done in those two hours.
     
  5. Jul 3, 2009 #4
    Re: limits

    Yea you both are right so that i can catch where my mistake is but it's so tough to write mathematical notations overhere.Could you plz give me a hint how can i write the mathematical notations?
     
  6. Jul 3, 2009 #5

    Office_Shredder

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  7. Jul 3, 2009 #6
    Re: limits

    thanks office_shredder i got the idea to use latex.

    Actually,for 1st question, i tried is a bit different one but similar to tht ones.The above questions seems complicated rather than i did.I gave yo cuz i get more idea to solve the question i tried.

    Ok the question is [similar to tht one]

    [tex]\lim_{x\rightarrow \infty} \sqrt{3x}-\sqrt{x-5}[/tex]

    = [tex]\lim_{x\rightarrow \infty} (\sqrt{3x}-\sqrt{x-5}) X (\sqrt{3x}+\sqrt{x-5}) [/tex]
     
  8. Jul 3, 2009 #7

    HallsofIvy

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    Re: limits

    No, you cannot just multiply by something without changing the value. You can multiply and divide by the same thing:
    [tex]\left(\sqrt{3x}- \sqrt{x-5}\right)\frac{\sqrt{3x}+ \sqrt{x- 5}}{\sqrt{3x}+ \sqrt{x- 5}}[/tex]
    [tex]= \frac{3x- (x- 5)}{\sqrt{3x}+ \sqrt{x- 5}}= \frac{2x+ 5}{\sqrt{3x}+ \sqrt{x- 5}}[/tex]

    Now, to take the limit as x goes to infinity, divide both numerator and denominator by x, remembering that x will become x2 inside the square roots.
     
  9. Jul 6, 2009 #8
    Re: limits

    Yea i know tht hallsofivy actually tht was my true mistake because i was just trying using latex & i can't get it well.

    actually.the process i'm gonna do is tht for the question
    [tex]
    (\sqrt{3x}-\sqrt{x-5})
    [/tex]

    is....
    I multiplied it by conjugate on both denominator and numerator sides & i got
    [tex]\frac{2x+5}{\sqrt{3x}+\sqrt{x-5}}[/tex]
    then i did 1st derivative on both numerator and denominator i got,

    [tex]\frac{4\sqrt{x(x-5)}}{\sqrt{3(x-5)}+\sqrt{x}}[/tex]

    Again on 2nd derivative I got,
    [tex]\frac{4x-10}{\sqrt{3x}+\sqrt{x-5}}[/tex]

    So,i can't converge this answer.Anyone have idea to solve this question.Where am i wrong?could yo plz point it out?
     
  10. Jul 6, 2009 #9
    Re: limits

    For the 2nd question i.e [tex]\lim_{x\rightarrow0}(tan{x}-sin{x})[/tex]
    1stly i find its derivative [on both sides numerator and denominator] i got,
    [tex]\frac{1-cos^3{x}}{3x^2cos^2{x}}[/tex]
    After further simplification with this i can't reach to the final answer.

    Again on next way i change the trigonometric terms into half angle rule using formulae either way i can't get it.
    Is there next way to solve this problem?
     
  11. Jul 7, 2009 #10
    Re: limits

    Since the denominator is a 3rd degree monomial, it is 0 as x approaches 0 up to the 3rd derivative, which should be a constant. Thus, you should look for the limit at the third application of L'Hopital's rule. Take care that the numerator remains 0 at each step as well, otherwise it would be invalid to apply L'Hopital's rule.
     
    Last edited: Jul 7, 2009
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