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I am trying to evaluate this limit for an assignment. $$\lim_{x \to \infty} \sqrt{x^2-6x +1}-x$$

I have tried to rationalize the function: $$=\lim_{x \to \infty} \frac{(\sqrt{x^2-6x +1}-x)(\sqrt{x^2-6x +1}+x)}{\sqrt{x^2-6x +1}+x}$$

$$=\lim_{x \to \infty} \frac{-6x+1}{\sqrt{x^2-6x +1}+x}$$

Then I multiply the function by $$\frac{(\frac{1}{x})}{(\frac{1}{x})}$$

Leading to

$$=\lim_{x \to \infty} \frac{-6+(\frac{1}{x})}{\sqrt{(\frac{-6}{x})+(\frac{1}{x^2})}+1}$$

Taking the limit, I see that all x terms tend to zero, leaving -6 as the answer. But -6 is not the answer. Why is that?

Flux
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4 Answers4

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You should have gotten, after the last step:

$$\lim_{x \to \infty} \frac{-6+\frac1x}{\sqrt{1-\frac6x +\frac1{x^2}}+1}=\frac{-6}{2}=-3$$

so in fact you only had a minor, though pretty influential, arithmetical mistake.

DonAntonio
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  • I'd say "algebraic" (not arithmetical) mistake. – Daniel R. Collins Oct 09 '16 at 19:35
  • @DanielR.Collins In fact, typo-related, as he simply forgot the $;\frac{x^2}{x^2}=1;$ summand under the square root sign. – DonAntonio Oct 09 '16 at 19:54
  • By direct calculation, when $0<y<1$ we have $(1-y/(2-y))^2<1-y<(1-y/2)^2, $ which is useful for problems like this. When $x>1,$ with $y=6/x-1/x^2,$ we have $x(1-y/(2-y))<\sqrt {x^2-6x+1}<x(1-y/2).$ – DanielWainfleet Oct 09 '16 at 21:36
  • @user254665 I don't really understand what your point is, partly because you didn't use /frac to write fractions and that way is hard to follow all that... – DonAntonio Oct 09 '16 at 22:47
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it should be $$\lim _{ x\to \infty } \frac { -6x+1 }{ \sqrt { x^{ 2 }-6x+1 } +x } =\lim _{ x\to \infty } \frac { x\left( -6+\frac { 1 }{ x } \right) }{ x\left( \sqrt { 1-\frac { 6 }{ x } +\frac { 1 }{ { x }^{ 2 } } } +1 \right) } =\frac { -6 }{ 2 } =-3$$

haqnatural
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Your error is here: $$\frac{\sqrt{x^2-6x +1}-x}{x}=\sqrt{1-\frac{6}{x}+\frac{1}{x^2}}+1$$

adjan
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It leads to

$$=\lim_{x \to \infty} \frac{-6+(\frac{1}{x})}{\sqrt{1-(\frac{6}{x})+(\frac{1}{x^2})}+1}$$

And so the limit is $-3$

marwalix
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