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Homework Help: Seeking help for limit math problem

  1. Sep 19, 2007 #1
    lim (sqrt(x^2+5) - 3)/(x-2)
    x->2


    i hope that kinda makes sense...

    anyways, my question is that ive been doing this and i keep getting 0 in the denominator in the answer...after subsituting everything i got 4/ sqrt(9) - 3 = 4/0
    so then is this equation just not possible?
     
    Last edited: Sep 19, 2007
  2. jcsd
  3. Sep 19, 2007 #2

    matt grime

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    How did you get this? That doesn't appear to bear any relation to the initial expression you gave.
     
  4. Sep 19, 2007 #3
    For [tex]\lim_{x \rightarrow 2}\frac{\sqrt{x^2+5}-3}{x-2}[/tex]

    Clearly the substitution method will give you a zero denominator!

    So, what methods have you learned to try and "avoid" a zero denominator?

    Casey

    I think they meant (sqrt(4+5)-3)/0....
    Arithmetic errors are always a good start:wink:
     
    Last edited: Sep 19, 2007
  5. Sep 19, 2007 #4
    um the course im taking told me first multiply the numerator and denominator by
    sqrt(x^2+5) - 3

    i might have done it wrong but then i got x^2+5-9/x-2(sqrt(x^2+5) - 3)
    i factored that to x^2-4/x-2(sqrt(x^2+5) - 3)
    and then eventually got (x-2)(x+2)/x-2(sqrt(x^2+5) - 3)
    canceled out x-2
    x+2/sqrt(x^2 + 5) - 3
    then subtituted the x for 2
    thats how i got that answer....i think thats how the teacher told me to do it...probably made mistakes along the way though..im not sure
     
  6. Sep 19, 2007 #5

    EnumaElish

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    Where is the interaction term in your numerator? How do you expand (a + b)^2?
     
  7. Sep 19, 2007 #6
    Your on the right track by using the conjugate; however, you need to watch your signs
    the conjugate of [tex]\sqrt{x^2+5}-3[/tex] is [tex]\sqrt{x^2+5}+3[/tex]

    Casey

    PS~Do you understand why the sign needs to be a plus (+) sign?
    Hint: [tex](a+b)^2 [/tex] does not = [tex]a^2+b^2[/tex]
    Casey
     
    Last edited: Sep 19, 2007
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