Limit of (x^2-6x+1)^(1/2)-x: Solve with Steps

[sinas]

limit as x->infinity of [(x^2-6x+1)^(1/2)-x]

I have tried to force it into a l'hopital form without much success, and tried to look up a couple different techniques (like replacing x with 1/u and finding the limit as u->zero) but I honestly don't even know where to begin.

 

[Pyrrhus]

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

First off L'Hospital rule only works if you get when evaluating your limit

\frac{\infty}{\infty} or \frac{0}{0}

Your limit evaluates to \infty - \infty

To solve it multiply the expression by \frac{\sqrt{x^2-6x+1} + x}{\sqrt{x^2-6x+1} + x}

 

[stunner5000pt]

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

First off L'Hospital rule only works if you get when evaluating your limit

\frac{\infty}{\infty} or \frac{0}{0}

Your limit evaluates to \infty - \infty

To solve it multiply the expression by \frac{\sqrt{x^2-6x+1} + x}{\sqrt{x^2-6x+1} + x}

And following from that, you get the expression

-6x + 1 / ((x^2-6x+1)+x)
divide the numerator into two fractions and understand that 1 / any value of x will go to zero eventually

also when you consider 6x watch how x^2 will increase so fast that it outpaces 6x +1 by leaps and bounds and you're left with simply x + x
voila!

 

[sinas]

Got it, thanks =)