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I am sorry if this is a stupid question, but I really have no clue how to find this limit.

$\lim \limits_{x \to \infty}({\sqrt{x^2-6x+8}-x})$ =$\lim \limits_{x \to \infty}\sqrt{x^2-6x+8} -\lim \limits_{x \to \infty}{x}$

It seems like $\infty-\infty$ case, and I do not know how to proceed.

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    $\begingroup$ For you to check, the answer is -3, you could multiply by the conjugate (see below) and use L'hospital's rule. $\endgroup$ – NoChance May 22 at 0:20
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Hint: It often helps to simplify the problem by a "conjugate" to make the square root disappear in the numerator. I.e. here, multiplying by $$ \frac{\sqrt{x^2-6x+8}+x}{\sqrt{x^2-6x+8}+x} $$ should help you see what's going on.

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    $\begingroup$ I cannot believe it was that easy. Thank you very much for your answer. $\endgroup$ – Radu Gabriel May 22 at 0:26
  • $\begingroup$ @RaduGabriel no problem, good luck! $\endgroup$ – qbert May 22 at 0:46
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$$x^2-6x+8=(x-3)^2-1$$

put

$$x-3=\cosh(t)$$

then

$$\sqrt{x^2-6x+8}-x=$$ $$\sinh(t)-\cosh(t)-3$$ $$=-e^{-t}-3$$

the limit when $t \to + \infty$ is $-3$.

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$\lim \limits_{x \to \infty}({\sqrt{x^2-6x+8}-x})$

rationalize. $$\frac{-6x+8} {\sqrt{x^2-6x+8}+x}$$ now take out x from denominator, so we are left with -6/2.

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