Why Can't I Solve This Limit Problem Without L'Hospital's Rule?

[redpepper007]

Homework Statement

[PLAIN]http://img545.imageshack.us/img545/5494/61141879.png

The answer is 2, but I just can't figure out how to get to it.

Homework Equations

none

The Attempt at a Solution

Tried all methods taught in school but none works. Adding related expression ((sqrt(16x^6-3x+4x^4)-2x^3) didnt help. Dividing everyhting by x^6 didnt help. Dividing by x^2 - nope... Tired to modify somehting, also no use... All the time I get either 0/0 or inf/inf... Dont know what to do else...

L'Hospital's method isn't allowed! ...

 

[Dick]

Divide numerator and denominator by x^3. And before you say "it doesn't work" show us what you got when you simplified.

 

[redpepper007]

Well, it works for sure, but I get different result...

http://img412.imageshack.us/img412/6621/mathq.th.jpg

And I believe 2 is the true answer, because I got the answer ''2'' from www.wolframalpha.com[/url] and [url]www.solvemymath.com[/URL] but they solve this using L'Hospitals method.. But I can't use it.

 

[jackmell]

Is that ink?

Also, you mess up the screen by posting something real wide like that, look, it's falling off the right side of the screen as I'm writing that. Finally, ain't that just 4/3+2/3? The whole thing is dominated by x^3. Those are the only terms that matter at infinity.

 

[redpepper007]

Yes, that's ink :D And fixed image, now nothing's messed up.

How do you tell that whole thing is dominated by x^3? The biggest x^ is x^6...

 

[Hurkyl]

The biggest x^ is x^6...

Ah, but it's wrapped inside an ( )^1/2, isn't it? That's why you divided the top and bottom by x³ rather than x⁶. (Where did the x⁹'s come from in your work?)

 

[redpepper007]

Ah, but it's wrapped inside an ( )^1/2, isn't it? That's why you divided the top and bottom by x³ rather than x⁶. (Where did the x⁹'s come from in your work?)

I got x⁹ when I divided sqrt(16x⁶-3x+4x⁴) with x³. And then I moved x³ under the sqrt sign, so x³ is ^2 so I get x⁹. (Because sqrt(x⁹)=x³)

 

[redpepper007]

I have this in my math notes:

''If x -> inf, the limit is equal to the ratio of biggest degrees of X, where each polynomial is of the same level.''

But I got confused because of that sqrt sign... I mean, does x⁶ and x⁴ also counts?

 

[jackmell]

Yes, that's ink :D And fixed image, now nothing's messed up.

How do you tell that whole thing is dominated by x^3? The biggest x^ is x^6...

Cus' I think writing in pencil should be part of the art of problem solving and if my professor insisted I do problems in ink then I'd say, "I'm sorry prof, but I'm gonna' have to drop your class cus' I'm not smart enough to never, ever make a math mistake and the scribble-scratch caused by scratching-out ink stuff just interferes with interpretation of an already difficult process."

I"m assuming you got that a term like $\sqrt{x^6}$ is asymptotic to $x^3$ for large x and all the other lower powers become insignificant for large x so that whole square root "looks" just like 4x^3 for large x.

 

[redpepper007]

I"m assuming you got that a term like $\sqrt{x^6}$ is asymptotic to $x^3$ for large x and all the other lower powers become insignificant for large x so that whole thing in the square root "looks" like just 4x^3 for large x.

umm, didnt understand what you meant with that. Can you use simpler English, please?

 

[Dick]

I got x⁹ when I divided sqrt(16x⁶-3x+4x⁴) with x³. And then I moved x³ under the sqrt sign, so x³ is ^2 so I get x⁹. (Because sqrt(x⁹)=x³)

When you move x^3 under the square root you get x^6. Because sqrt(x^6)=x^3. Not x^9.

 

[redpepper007]

When you move x^3 under the square root you get x^6. Because sqrt(x^6)=x^3. Not x^9.

But how can that be true? sqrt(9)=3 but sqrt(6)=2.44948...

 

[jackmell]

umm, didnt understand what you meant with that. Can you use simpler English, please?

Ok, I made a mistake with that: $\sqrt{x^6}$ IS x^3. I mean things like $\sqrt{P_6(x)}$ where P_6(x) is a sixth-degree polynomial. For very large x, all the lower-power x terms contribute little to the entire value of the polynomial because the x^6 term is much, much larger than all the other terms no matter how large the coefficients on the other powers are. So we can write:

$$\lim_{x\to\infty} P_6(x)\sim a x^6$$

meaning that as x grows very large, the polynomial becomes closer and closer to just the x^6 term.

 

[redpepper007]

umm, is this like that:

sqrt(x^6)=((x^6)^0.5)=x^6*0.5=x^3 ?

Oh, I think I got it... wait a few minutes, will post the result!

http://img718.imageshack.us/img718/5571/math2z.th.jpg YAYAYAYAYAAA! :p Thanks a lot!

 

[jackmell]

$$\lim_{x\to\infty}\frac{\sqrt{16x^6-3x+4x^4)}+2x^3}{3x^3+2x^2+3}$$

Now divide top and bottom by:

$$x^3=\sqrt{x^6}$$:

$$\frac{\sqrt{\frac{16x^6-3x+4x^4}{x^6}}+\frac{2x^3}{x^3}}{\frac{3x^3+2x^2+3}{x^3}}$$

Now try and just look at that expression and imagine what happens as x gets very large? The root argument tends to 16 right? that other term in the numerator tends to 2 and the bottom tends to 3.

Edit: see you got it but I'll leave my up as well. Also, looks like the ink looks better compared to what you posted in pencil but still I'd recommend you use pencil for math.