Limit proof as x approaches infinity

[fishturtle1]

Homework Statement

Verify the following assertions:
a) ##x^2 + \sqrt{x} = O(x^2)##

2. Homework Equations
If the limit as x approaches ##\infty## of ##\frac {f(x)}{g(x)}## exists (and is finite), then ##f(x) = O(g(x))##.

The Attempt at a Solution

Let ##\epsilon > 0##. We solve for ##\delta## such that ##x > \delta## implies ##|\frac{f(x)}{g(x)} - 1| < \epsilon##. Consider ##|\frac {x^2 + \sqrt{x}}{x^2} - 1| < \epsilon##. Then ##\frac {x^2 + \sqrt{x}}{x^2} - 1 < \epsilon##, so ##\frac {x^2 + \sqrt{x}}{x^2} < \epsilon + 1##. Taking the inverse of both sides and then taking the square root of both sides, we get ##\frac x{\sqrt{\sqrt{x} + x^2}} > \frac 1{\sqrt{\epsilon + 1}}##..I'm trying to get to an expression "x > some expression in terms of ##\epsilon##" and then substitute that for ##\delta##.

 

[mfb]

Why would you take the inverse square root of both sides?
$$\frac {x^2 + \sqrt{x}}{x^2} < \epsilon + 1$$That is good. Now simplify it.

 

[fishturtle1]

Why would you take the inverse square root of both sides?
$$\frac {x^2 + \sqrt{x}}{x^2} < \epsilon + 1$$That is good. Now simplify it.

Thanks for the response,

##\frac {x^2 + \sqrt{x}}{x^2} = 1 + \frac{\sqrt{x}}{{x^2}} = 1 + \frac 1{x^{3/2}} < \epsilon + 1## so, ##\frac 1{x^{3/2}} < \epsilon##
Raising both sides to ##-5/2##, we get ##x > \epsilon^{\frac{-5}{2}} = \delta##. Thus, we can conclude the limit as x approaches ##\infty## for ##\frac {x^2 + \sqrt{x}}{x^2} = 1##. Thus, ## {x^2 + \sqrt{x}} = O(x^2)##.

 

[Ray Vickson]

Homework Statement

Verify the following assertions:
a) ##x^2 + \sqrt{x} = O(x^2)##

2. Homework Equations
If the limit as x approaches ##\infty## of ##\frac {f(x)}{g(x)}## exists (and is finite), then ##f(x) = O(g(x))##.

The Attempt at a Solution

Let ##\epsilon > 0##. We solve for ##\delta## such that ##x > \delta## implies ##|\frac{f(x)}{g(x)} - 1| < \epsilon##. Consider ##|\frac {x^2 + \sqrt{x}}{x^2} - 1| < \epsilon##. Then ##\frac {x^2 + \sqrt{x}}{x^2} - 1 < \epsilon##, so ##\frac {x^2 + \sqrt{x}}{x^2} < \epsilon + 1##. Taking the inverse of both sides and then taking the square root of both sides, we get ##\frac x{\sqrt{\sqrt{x} + x^2}} > \frac 1{\sqrt{\epsilon + 1}}##..I'm trying to get to an expression "x > some expression in terms of ##\epsilon##" and then substitute that for ##\delta##.

You have
$$\frac{x^2+\sqrt{x}}{x^2} = 1 + \frac{1}{x^{3/2}} \leq 2$$
for ##x \geq 1##. Thus, ##x^2 + \sqrt{x} \leq k x^2## where ##k = 2##. If you further restrict the range of ##x > 0## you can make ##k## smaller, but that is not relevant to the definition of ##O(x^2)##, which just need some finite ##k > 0##.

BTW: your definition of ##O## is different from those I have seen; the ones I know of do not require that ##f(x)/g(x)## have a finite limit as ##x \to \infty##, they just require that there exists some finite ##k > 0## such that ##|f(x)/g(x)| \leq k## as ##x \to \infty##.

 

[fishturtle1]

You have
$$\frac{x^2+\sqrt{x}}{x^2} = 1 + \frac{1}{x^{3/2}} \leq 2$$
for ##x \geq 1##. Thus, ##x^2 + \sqrt{x} \leq k x^2## where ##k = 2##. If you further restrict the range of ##x > 0## you can make ##k## smaller, but that is not relevant to the definition of ##O(x^2)##, which just need some finite ##k > 0##.

BTW: your definition of ##O## is different from those I have seen; the ones I know of do not require that ##f(x)/g(x)## have a finite limit as ##x \to \infty##, they just require that there exists some finite ##k > 0## such that ##|f(x)/g(x)| \leq k## as ##x \to \infty##.

Thanks for the clarification, I didn't know about the actual definition of ##O##.

I should add that the textbook says this method can *sometimes* be used to prove that ##f(x) = O(g(x))##. So I guess the book's method only works for certain situations.. and even if it fails, it doesn't rule out whether or not ##f(x) = O(g(x))##.

 

[mfb]

You showed more than just O(x²), but showing that the limit is 1 shows the existence of the limit, of course.