What Is the Correct Approach to Evaluate Limits at Infinity with Square Roots?

[credico]

Homework Statement

Evaluate the following limits:

lim sqrt(x^2-3x+1)-x
x->\infty


lim sqrt(x^2-3x+1)-x
x->-\infty

2. The attempt at a solution

http://img816.imageshack.us/img816/9995/limitproblem11.jpg

We have to enter in the answers online into a program that tells us if we're right or wrong. I got the limit going to positive infinity correct, but I don't know what I am doing wrong going to negative infinity.

 

[hunt_mat]

Take out a factor of x^2 from the square root and expand \sqrt{1+X} as a power series and that should solve the problem for you.

 

[SammyS]

Homework Statement

Evaluate the following limits:

lim_x→-∞ sqrt(x^2-3x+1)-x


2. The attempt at a solution

We have to enter in the answers online into a program that tells us if we're right or wrong. I got the limit going to positive infinity correct, but I don't know what I am doing wrong going to negative infinity.

You did a fine job recognizing that \frac{1}{x}=-\,\frac{1}{\sqrt{x^2}} for x < 0.

Once you get to \lim_{x\to -\infty}\left(\sqrt{x^2-3x+1}-x\right)=\lim_{x\to -\infty}\left(\frac{-3x+1}{x+\sqrt{x^2-3x+1}}\right)\,, try using L'Hôpital's rule.

Added during Edit:

Wait! Look at the original! \lim_{x\to -\infty}\left(\sqrt{x^2-3x+1}-x\right).

\lim_{x\to -\infty}\left(\sqrt{x^2-3x+1}\right)=+\infty

and

\lim_{x\to -\infty}\left(-x\right)=+\infty

 

[credico]

You did a fine job recognizing that \frac{1}{x}=-\,\frac{1}{\sqrt{x^2}} for x < 0.

Once you get to \lim_{x\to -\infty}\left(\sqrt{x^2-3x+1}-x\right)=\lim_{x\to -\infty}\left(\frac{-3x+1}{x+\sqrt{x^2-3x+1}}\right)\,, try using L'Hôpital's rule.

Added during Edit:

Wait! Look at the original! \lim_{x\to -\infty}\left(\sqrt{x^2-3x+1}-x\right).

\lim_{x\to -\infty}\left(\sqrt{x^2-3x+1}\right)=+\infty

and

\lim_{x\to -\infty}\left(-x\right)=+\infty

Haven't learned L'Hopital's rule yet (I know what it is but at this point I think they want us to do it a certain way first).

In regards to what you posted in your edit: You used the Limit Laws to separate the two, but how did you calculate the radical limit? Just by plugging in infinity (or looking at what it does near infinity?) Is that allowed? Are you certain there isn't something that can mislead me by doing this, at least in this case?

Thanks.

Your answer was right by the way. I guess if it's continuous, by definition: lim f(x) as x ->a = lim f(a). It just seems a bit odd since there's that radical floating over top.

 

[HallsofIvy]

You did a fine job recognizing that \frac{1}{x}=-\,\frac{1}{\sqrt{x^2}} for x < 0.

Once you get to \lim_{x\to -\infty}\left(\sqrt{x^2-3x+1}-x\right)=\lim_{x\to -\infty}\left(\frac{-3x+1}{x+\sqrt{x^2-3x+1}}\right)\,, try using L'Hôpital's rule.

Better: Divide both numerator and denominator by x to get
\frac{-3+ \frac{1}{x}}{\sqrt{1- \frac{3}{x}+ \frac{1}{x^2}}}
Of course, as x goes to infinity, 1/x or 1/x^2 goes to 0. For the case were x is going to negative infinity, divide by -x (which will be positive) so that you have no problem with it inside the square root.

Added during Edit:

Wait! Look at the original! \lim_{x\to -\infty}\left(\sqrt{x^2-3x+1}-x\right).

\lim_{x\to -\infty}\left(\sqrt{x^2-3x+1}\right)=+\infty

and

\lim_{x\to -\infty}\left(-x\right)=+\infty