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Finding Constants To A Limit Problem

  1. May 31, 2012 #1
    Find all the values of the constants a and b such that
    [itex]\lim_{x\rightarrow0}\frac{\sqrt{a+bx} -\sqrt{3}}{x} = \sqrt{3}[/itex]

    I tried to multiply by the conjugate, but it indeed failed to give me any insight as to how to solve this problem. Could someone possibly prod me into the right direction?
     
  2. jcsd
  3. May 31, 2012 #2

    tiny-tim

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    Hi Bashyboy! :smile:

    Start with a …

    what must a be?​

    (can a = 0? 2? …)
     
  4. May 31, 2012 #3
    So, in order to start this problem, we have to begin with speculation? There is no analytic way of solving for a? Well, I know it can't be zero, because then all you would have under the radical is zero.
     
  5. May 31, 2012 #4

    SammyS

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    Since the denominator → 0 , the only way for the limit to exist is for the numerator → 0 as x → 0. Right ?
     
  6. May 31, 2012 #5

    tiny-tim

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    No, but trying one or two values may help us to see what we need to avoid :wink:
     
  7. May 31, 2012 #6
    Note that this is an equation with a limit on one side. The limit, if it converges evaluates to a number. So this is a relationship between two numbers.

    Does the statement as provided imply the validity of any further statements? Equivalent ones that can be manipulated and better understood to reach an answer?

    Consider that the given equation implies:

    [itex]\frac{\lim_{x \to 0}\sqrt{a+bx}-\sqrt{3}}{\lim_{x \to 0}x} = \sqrt{3}[/itex]

    which implies:

    [itex]\lim_{x \to 0}\sqrt{a+bx}-\sqrt{3} = \lim_{x \to 0}\sqrt{3}x[/itex]

    From this, you should be able to figure out certain restrictions on the variables [itex]a[/itex] and [itex]b[/itex].
     
  8. May 31, 2012 #7

    SammyS

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    It looks like it will be a valid two-sided limit when all is done.
    What you have written above is absolutely improper. The denominator you show is identically zero.
     
  9. Jun 1, 2012 #8
    Okay, so I set the numerator equal to zero, and took the limit of it, and I got a = 3; but now I am not sure where to plug in a = 3 so I can solve for b.
     
  10. Jun 1, 2012 #9
    Since you have both your numerator and denominator approaching zero, that means you have a 0/0 indeterminate limit form. How do you solve such limits?
     
  11. Jun 1, 2012 #10

    SammyS

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    Now that you have a =3, look at your result from multiplying by the conjugate. That should tell you what to use for b.
     
  12. Jun 1, 2012 #11
    Does it sound right that b = 6?
     
  13. Jun 1, 2012 #12
    Yep!! That is correct :approve:
     
  14. Jun 1, 2012 #13
    Thank you all very much, all of you provided great insight.
     
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