• credico
In summary: 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:Better: Divide both numerator and denominator by x to get:\frac{

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 [Broken]

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.

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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.

credico said:

Homework Statement

Evaluate the following limits:

limx→-∞ 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.

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

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

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SammyS said:
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.

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.

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SammyS said:
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.

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

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

What is a limit?

A limit is a mathematical concept that describes the behavior of a function as the input values approach a certain value. It is represented by the symbol "lim" and is used to understand the behavior of functions near certain points or values.

What is infinity?

Infinity is a concept in mathematics that represents a quantity that is larger than any real number. It is often denoted by the symbol "∞" and is used to describe unbounded or endless quantities.

What is a radical?

A radical is a mathematical symbol that represents a root of a number. It is denoted by the symbol "√" and is used to find the value that, when multiplied by itself a certain number of times, equals the given number.

How are limits, infinity, and radicals related?

Limits, infinity, and radicals are all concepts that are commonly used in calculus and other branches of mathematics. Limits can be used to understand the behavior of functions near certain values, infinity can be used to describe unbounded quantities, and radicals can be used to find the roots of numbers. These concepts are often used together to solve complex mathematical problems.

What are some real-world applications of limits, infinity, and radicals?

Limits, infinity, and radicals have many real-world applications, such as in physics, engineering, and economics. For example, limits can be used to understand the velocity of an object as it approaches a certain point, infinity can be used to model population growth, and radicals can be used in calculating interest rates for loans.