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Evalute the following limit

  1. Nov 16, 2013 #1
    Hey guys, I have been trying this question for 40 minutes and I can't get the answer.
    Evaluate lim as x approaches positive infinity(sqrt(x^2+1)-x).
    I tried the La-Hopital's rule, but even then I still get inf/inf. I know there is an algebraic manipulation I need to make, but I keep getting it wrong.
    Thanks in advance.
     
  2. jcsd
  3. Nov 16, 2013 #2

    hilbert2

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    I'd suggest you show that if ##f(x)=\sqrt{x^{2}+1}-x##, then ##0 \leq f(x) \leq \sqrt{x^{2}}+\frac{1}{x}-x## for positive ##x##, and because ##\lim_{x \to \infty}\left(\sqrt{x^{2}}+\frac{1}{x}-x\right)=0##, the limit of ##f(x)## must be zero when ##x## approaches infinity.

    You can prove that ##f(x) \leq \sqrt{x^{2}}+\frac{1}{x}-x## by assuming the contrary and deriving a contradiction.
     
  4. Nov 16, 2013 #3
    Thanks hilbert2,
    I see what you are saying. I also just found another way to do it, ill post it:
    Multiplying the expression by (sqrt(x^2+1) + x)/(sqrt(x^2+1)+x) you get lim(1/(sqrt(x^2+1)+x) = 0.
    I see that you have used the sandwich theorem, and i suppose that is an easier way to do it. I am confused on how you got your upper limit 1/x, could you please explain how you got that upper limit. Thanks
     
  5. Nov 16, 2013 #4
    Ah, did you just arbitrarily pick the upper limit which is larger than f(x)?
     
  6. Nov 16, 2013 #5

    hilbert2

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    Let's suppose the contrary, that ##\sqrt{x^{2}+1}>\sqrt{x^{2}}+\frac{1}{x}##. As we can suppose ##x## is positive, we can square both sides of the inequality to get
    ##x^{2}+1>x^{2}+2\frac{|x|}{x}+\frac{1}{x^{2}}##. When ##x## is positive, ##\frac{|x|}{x}=1##, and therefore we get ##1>2+\frac{1}{x^{2}}##, which is a contradiction.

    You can choose any upper limit that approaches zero when ##x \to \infty##.
     
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