# Square roots and square

1. Jun 17, 2007

### lLovePhysics

I don't get this problem and why the answer is what the book states that it should be:

If $$f(x)=\sqrt{x^{2}}$$ then f(x) can also be expressed as: l x l

The answer I chose was simply x , but I don't know why it is wrong.

2. Jun 17, 2007

### Hurkyl

Staff Emeritus
From a practical point of view, you memorized a wrong statement
$x = \sqrt{x^2}$​
and you need to unlearn it and memorize the correct statement
$|x| = \sqrt{x^2}$ (for any real number x).​

As for "why" it's wrong, at what values of x does your answer disagree with the correct answer? Have you tried plugging one of those values into the expression $\sqrt{x^2}$?

3. Jun 17, 2007

### lLovePhysics

$$\lim_{x\rightarrow\infty} (\frac{3x^{2}+4x-5}{6x^{2}+3x+1})$$

How do you solve this algebraically/numerically? I've tried graphing it on my calculator and I can clearly see that it approached .5 but how come when I put infinity into x in the table it f(x) shows up as 1?? Weird..

4. Jun 17, 2007

### Hurkyl

Staff Emeritus
Table? What table?

This is one of the standard examples in your calculus book: when you have rational functions (or rational function-like things), and you want to analyze them "at" infinity, you should divide the numerator and denominator by the highest power of your variable that appears in the expression.

5. Jun 17, 2007

### HallsofIvy

Staff Emeritus
Already answered but, in particular, suppose x= -3. Then x2= 9. $\sqrt{9}= 3$, not -3 so $\sqrt{x^2}$ is NOT x!

You DON'T "put infinity into x"- the function is only defined for x real and "infinity" is NOT a real number. Since we know that $\lim_{x\rightarrow \infty} 1/x= 0$, try dividing both numerator and denominator by x2: we now have
$$\lim_{x\rightarrow\infty}\frac{3+ \frac{4}{x}-\frac{5}{x^2}}{6+ \frac{3}{x}+ \frac{1}{x^2}}$$
For x very very large, those fractions in the numerator and denominator are very very small- they go to 0. So what does the whole thing go to?

6. Jun 17, 2007

### lLovePhysics

Oh, I'm not in calculus yet so I didn't know. So what if the expression was $$\frac{x^{3}}{x^{2}}$$, you would have to divide both numerator and denominator by x^3?

7. Jun 17, 2007

### lLovePhysics

Okay I see, you divided by x^2 so that you can make some numbers equal zero. 3/6= 1/2

8. Jun 17, 2007

### Hurkyl

Staff Emeritus
Yep. And then since the numerator goes to 1 and the denominator goes to 0, you learn that this fraction has no limit!

9. Jun 17, 2007

### lLovePhysics

Ooo, okay I get it now. Thanks guys

10. Jun 17, 2007

### ice109

ehh it should really be done using l'hospitale but you're not there yet

11. Jun 17, 2007

### Hurkyl

Staff Emeritus
Why do you think that? The method I described is perfectly rigorous.

12. Jun 17, 2007

### symbolipoint

Just a comment: about that polynomial rational limit expression, you will learn about these in SOME Precalculus-Elementary Functions courses, but not in others -------- depends where you attend.

13. Jun 17, 2007

### robert Ihnot

Originally Posted by ice109
ehh it should really be done using l'hospitale but you're not there yet

Hurky: Why do you think that? The method I described is perfectly rigorous.

Of course, and another way to look at it is to realize that the leading term will dominate in a polynominal, so all that matters is 3X^2/6X^2, which goes to 1/2. This may not appear very rigorous, but it can be seen immediately.

14. Jun 17, 2007

### ice109

honestly i don't know the rigor behind the division, it seems a little arbitrary. though i do think you could probably tell me.

15. Jun 18, 2007

### chroot

Staff Emeritus
What's not rigorous about multiplying the top and bottom of a fraction by the same thing?

You're just multiplying the fraction by, e.g.

$\frac{1/x^2}{1/x^2}$

- Warren