- #1

lLovePhysics

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If [tex]f(x)=\sqrt{x^{2}}[/tex] 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.

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- Thread starter lLovePhysics
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- #1

lLovePhysics

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If [tex]f(x)=\sqrt{x^{2}}[/tex] 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

Hurkyl

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[itex]x = \sqrt{x^2}[/itex]

and you need to unlearn it and memorize the correct statement[itex]|x| = \sqrt{x^2}[/itex] (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 [itex]\sqrt{x^2}[/itex]?

- #3

lLovePhysics

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[tex]\lim_{x\rightarrow\infty} (\frac{3x^{2}+4x-5}{6x^{2}+3x+1})[/tex]

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

Hurkyl

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Table? What table?

[tex]\lim_{x\rightarrow\infty} (\frac{3x^{2}+4x-5}{6x^{2}+3x+1})[/tex]

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..

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

HallsofIvy

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Already answered but, in particular, suppose x= -3. Then x

If [tex]f(x)=\sqrt{x^{2}}[/tex] 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.

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 [itex]\lim_{x\rightarrow \infty} 1/x= 0[/itex], try dividing both numerator and denominator by x

[tex]\lim_{x\rightarrow\infty} (\frac{3x^{2}+4x-5}{6x^{2}+3x+1})[/tex]

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..

[tex]\lim_{x\rightarrow\infty}\frac{3+ \frac{4}{x}-\frac{5}{x^2}}{6+ \frac{3}{x}+ \frac{1}{x^2}}[/tex]

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

lLovePhysics

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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.

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

- #7

lLovePhysics

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Already answered but, in particular, suppose x= -3. Then x^{2}= 9. [itex]\sqrt{9}= 3[/itex], not -3 so [itex]\sqrt{x^2}[/itex] 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 [itex]\lim_{x\rightarrow \infty} 1/x= 0[/itex], try dividing both numerator and denominator by x^{2}: we now have

[tex]\lim_{x\rightarrow\infty}\frac{3+ \frac{4}{x}-\frac{5}{x^2}}{6+ \frac{3}{x}+ \frac{1}{x^2}}[/tex]

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?

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

- #8

Hurkyl

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Yep. And then since the numerator goes to 1 and the denominator goes to 0, you learn that this fraction has no limit!Oh, I'm not in calculus yet so I didn't know. So what if the expression was [tex]\frac{x^{3}}{x^{2}}[/tex], you would have to divide both numerator and denominator by x^3?

- #9

lLovePhysics

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

Ooo, okay I get it now. Thanks guys

- #10

ice109

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

- #11

Hurkyl

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Why do you think that? The method I described is perfectly rigorous.ehh it should really be done using l'hospitale but you're not there yet

- #12

symbolipoint

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- #13

robert Ihnot

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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

ice109

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

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

chroot

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

[itex]\frac{1/x^2}{1/x^2}[/itex]

- Warren

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