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?
Best Answer
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.