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Limit problem with square root

  1. Aug 22, 2015 #1
    I have done this problem before but forgot how to get from one step to the next:

    let a>0.

    how is absval(x^1/2-a^1/2) equal to abval(x-a)/(x^1/2-a^1/2)?
     
  2. jcsd
  3. Aug 22, 2015 #2
    I think you want to express
    $$\sqrt{x}-\sqrt{a}=\frac{(\sqrt{x}-\sqrt{a})(\sqrt{x}+\sqrt{a})}{\sqrt{x}+\sqrt{a}}=\frac{x-a}{\sqrt{x}+\sqrt{a}}$$
    right? If so, it just applies the law ##(a+b)(a-b)=a^2-b^2.##
     
  4. Aug 22, 2015 #3
    They are not equal. What was the original question?
     
  5. Aug 22, 2015 #4
    but the abs val goes away on the denom. that is what I don't understand, you can't assume root x - root a is positive
     
  6. Aug 22, 2015 #5
    let f(x) equal root x show that as x→a lim f(x) equals root a
     
  7. Aug 22, 2015 #6
    also a>0
     
  8. Aug 22, 2015 #7
    sorry I meant you can't assume root x plus root a is positive, if say c∈ℝ and c>0 then abs val(c⋅x)=c⋅abs val(x)
     
  9. Aug 22, 2015 #8
    never mind I figured out, but thanks for the replies
     
  10. Aug 22, 2015 #9
    To show ##\displaystyle{\lim_{x\rightarrow a}}\sqrt{x}=\sqrt{a},## you ought to prove for any ##x\in (a-\delta, a+\delta),## ##\exists \varepsilon## such that ##|f(x)-f(a)|<\varepsilon.##
     
  11. Aug 22, 2015 #10
    Thanks, yeah I have done this proof before. I just made a mistake thinking about negatives when clearly root x plus root a is always positive and therefore can be pulled out of the abs val
     
  12. Aug 22, 2015 #11
    I mean it assumes that x>0. The problem did not state that although I guess it is just assumed considering it is real analysis.
     
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