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TWJian

Limits at infinity and the square root of infinity

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I know that infinity is indefinite, and thus could be regarded as not being a number. With that, I realize that attempting to perform any arithmetic operations on infinity(other then 1/infinity, but that technically isn't an arithmetic operation either, though it looks like one) is both futile and meaningless.

 

I thought that the limit of f(x)=x^(0.5) or (x+(x+4)^0.5)^0.5 is infinity as x approaches infinity. However, the graph of y= x^0.5 appears to have an horizontal asymptote (it does however, appear to be indefinite) Why is there such an asymptote? Is the square root of infinity some other indefinite constant which is not infinity? Or is the square root still infinity, though it is part of a different, yet still infinite set? Is the numerical value of the square root of infinity, infinity if we try to assign a value to it? Or is the square root of infinity impossible to define in terms of infinity (since both are indefinite), and thus could only be expressed as itself, the square root of infinity? (surd infinity is surd infinity)

 

Incidentally, does a limit exist if the function approaches infinity as x approaches infinity? Since there is no actual boundry, a limit, apparently does not exist (to me). Furthermore, the limit appears to be one-sided, since it seems impossible to approach infinity from the right (or negative infinity from the left, for that matter)

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I thought that the limit of f(x)=x^(0.5) or (x+(x+4)^0.5)^0.5 is infinity as x approaches infinity.

It is.

 

However, the graph of y= x^0.5 appears to have an horizontal asymptote (it does however, appear to be indefinite) Why is there such an asymptote?

There is no such asymptote, but the rise of the graph goes to zero: [math] \lim _{x \rightarrow \infty} \frac{1}{2} x^{\frac{-1}{2}} = 0[/math]

 

Is the square root of infinity some other indefinite constant which is not infinity?

[math]\lim _{x \rightarrow \infty} \sqrt{x} = \infty [/math]

 

Or is the square root still infinity, though it is part of a different, yet still infinite set? Is the numerical value of the square root of infinity, infinity if we try to assign a value to it? Or is the square root of infinity impossible to define in terms of infinity (since both are indefinite), and thus could only be expressed as itself, the square root of infinity? (surd infinity is surd infinity)

I don't understand that.

 

Incidentally, does a limit exist if the function approaches infinity as x approaches infinity?

No for real-valued limits. Infinity means larger than any chosen value. That includes larger than any chosen limit (unless you accept "infinity" as a result for the limit, in which case the answer is obvious).

 

Since there is no actual boundry, a limit, apparently does not exist (to me).

Saying the limit is infinity actually kind-of (don't want to think about or look up a formal definition) means that the function grows beyond every boundry, so in some sense your interpretation is correct. You might see the limit being infinity as a special case of that a limit does not exsits. Note, however, that this is not an equivalence statement. The limit of sin(x) for x to infinity does not exist either. But that function does not grow beyond all borders.

 

Furthermore, the limit appears to be one-sided, since it seems impossible to approach infinity from the right (or negative infinity from the left, for that matter)

Don't understand that, either.

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I thought that the limit of f(x)=x^(0.5) or (x+(x+4)^0.5)^0.5 is infinity as x approaches infinity. However, the graph of y= x^0.5 appears to have an horizontal asymptote (it does however, appear to be indefinite) Why is there such an asymptote?

 

Short answer: there isn't. Unfortunately when we look at the graphs of functions, it sometimes leads us to infer statements which are not true. That's why it's essential to impose logic and rigour to ensure that the results we prove are correct. The truth of the matter is that [imath]\sqrt{x}[/imath] does tend to infinity, but it does it extremely slowly.

 

Here's the proof. We want to show that [imath]f(x) = \sqrt{x}[/imath] is unbounded. What does that mean? Simply, that for any possible value [imath]M \in \mathbb{R}[/imath], there exists another value [imath]m \in \mathbb{R}[/imath] such that [imath]\sqrt{m} > M[/imath].

 

So, given [imath]M[/imath], we pick any [imath]m > M^2[/imath] - this is obviously possible. Then by taking square roots of the inequality, we get [imath]\sqrt{m} = f(x) > |M|[/imath]. Hence f is unbounded, and since f is strictly increasing ([imath]a < b \Rightarrow f(a) < f(b)[/imath]), we can conclude that [imath]\lim_{x\to\infty} f(x) = \infty[/imath].

 

That is a rock solid argument - sometimes you should not believe what your eyes are saying!

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In plain English, I could pick any positive number of y, no matter how large, and get that value by plugging in a large enough value into the square root of x. Try it.

 

For example, when f(x) = 10^100, x must be 10^200.

 

When f(x) = 10^10,000, x must be 10^20,000.

 

Any positive number can result from the square root of x. The exponent must simply be twice as large.

 

If it was a horizontal asymptote, you couldn't get any result, only those below it.

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