Question

Evaluate the limit as x approaches positive infinity:

sqrt(X^2 - 6x +1)-x

Answer 1

\[\lim_{x \rightarrow \infty} \sqrt{x^{2}-6x+1}-x\]

Answer 2

answer is -3 and i can show at least the first step

Answer 3

multiply top and bottom as if you were rationalizing the denominator by
\[\sqrt{x^2-6x+1}+x\]
\[\frac{x^2-6x+1-x^2}{\sqrt{x^2-6x+1}+x}\]
combine like terms in the numerator to get
\[\frac{-6x+1}{\sqrt{x^2-6x+1}+x}\]
and now let x go to infinity. you will get
\[\frac{-6}{2}=-3\] but the details for that step are up to you. basically the only terms that matter in the denominator are
\[\sqrt{x^2}\] and
\[x\]

Answer 4

i have seen this written out where you divide each term by x. that may be a good explanation, with the sophistication that if you divide by x inside the radical you you have to write it as
\[\frac{1}{x^2}\]so the denominator will look something like
\[\sqrt{\frac{x^2}{x^2}-\frac{6x}{x^2}+\frac{1}{x^2}}+\frac{x}{x}\] or
\[\sqrt{1-\frac{6}{x}+\frac{1}{x^2}}+1\] and the numerator will look like
\[-\frac{6x}{x}+\frac{1}{x}=-6+\frac{1}{x}\]

Answer 5

then let x go to infinity, argue that each term with an "x" in the denominator goes to zero, and you are left with
\[-\frac{6}{2}\]