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

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.

• For you to check, the answer is -3, you could multiply by the conjugate (see below) and use L'hospital's rule. – NoChance May 22 at 0:20

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.

• I cannot believe it was that easy. Thank you very much for your answer. – Radu Gabriel May 22 at 0:26
• @RaduGabriel no problem, good luck! – qbert May 22 at 0:46

$$x^2-6x+8=(x-3)^2-1$$

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 $$-3$$.

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

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