# Homework Help: I just cant solve this specific limit

1. Sep 18, 2010

### redpepper007

1. The problem statement, all variables and given/known data

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

The answer is 2, but I just cant figure out how to get to it.

2. Relevant equations

none

3. 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 isnt allowed!!! ...

Last edited by a moderator: May 4, 2017
2. Sep 18, 2010

### Dick

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

3. Sep 18, 2010

### redpepper007

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

http://img412.imageshack.us/img412/6621/mathq.th.jpg [Broken]

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 cant use it.

Last edited by a moderator: May 4, 2017
4. Sep 18, 2010

### 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.

5. Sep 18, 2010

### redpepper007

Yes, thats 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...

6. Sep 18, 2010

### Hurkyl

Staff Emeritus
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?)

7. Sep 18, 2010

### redpepper007

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)

8. Sep 18, 2010

### 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?

9. Sep 18, 2010

### jackmell

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.

Last edited: Sep 18, 2010
10. Sep 18, 2010

### redpepper007

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

11. Sep 18, 2010

### Dick

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

12. Sep 18, 2010

### redpepper007

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

13. Sep 18, 2010

### jackmell

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.

14. Sep 18, 2010

### redpepper007

Last edited by a moderator: May 4, 2017
15. Sep 18, 2010

### 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.