What is the limit of the square root of a polynomial as x approaches infinity?

[kitle545]

Homework Statement

lim (SQRT (8x^3 + 5x + 10) ) / x^2

x -> infinity

Homework Equations

The Attempt at a Solution

I tried factoring out x^3, but that didn't help anything.

 

[cse63146]

\frac{A+B+C}{D} = \frac{A}{D} + \frac{B}{D} + \frac{C}{D}

 

[kitle545]

The square root sign is over the (8x^3 + 5x + 10)

 

[boombaby]

= lim SQRT ((8x^3 + 5x + 10) / x^4)
does this help?

 

[JG89]

1 < x^5/8 - 5/8x^2 is true for large values of x (notice that the second term goes to 0 x approaches infinite). So we then have:


8x^2 + 5 < x^7

And because of that, we have (because the term 10/x goes to 0 as x approaches infinite):
8x^2 + 10/x + 5 < x^7

x^2 < x^7/8 - 10/8x - 5/8

8x^2 < x^7 - 10/x - 5

8x^2 + 5 < x^7 - 10/x

x(8x^2 + 5) < x^8 - 10

8x^3 + 5x + 10 < x^8

SQRT ((8x^3 + 5x + 10)< x^4



So, we see that the denominator goes to infinite faster than the numerator. What is the limit then?

 

[boombaby]

If you can get f(x)=(8x^3 + 5x + 10) / x^4 ->0, as x->+\infty with no problem, then Sqrt(f(x))-> Sqrt(0), as x->+\infty
In this case you may want to show that if f(x)>=0 and f(x)->L, then Sqrt(f(x))->Sqrt(L), as x->p , where p is a limit point of it's domain.

There should be many other ways to explain the limit, based on the continuity of sqrt(x) maybe.