Solving Infinite Limit: \frac {t}{\sqrt{4t^{2}+1}}

[Asphyxiated]

Homework Statement

$$\left ^{lim} _{t\rightarrow \infty} \right \frac {t}{\sqrt{4t^{2}+1}}$$

Homework Equations

The Attempt at a Solution

Not to confident on Limits but I can usually get them, I just don't know how to get all the way there:

Let:

$$H = \infty$$

$$\frac {H}{\sqrt{4H^{2}+1}}$$

I don't know what to do from here to actually get a numerical result, but to me it looks like it would basically be H/sqrt(H^2) and should be one or zero but the actual answer they are looking for is 1/2 so could someone please instruct me on how to get there?

thanks!

 

[Mark44]

Homework Statement

$$\left ^{lim} _{t\rightarrow \infty} \right \frac {t}{\sqrt{4t^{2}+1}}$$

Homework Equations

The Attempt at a Solution

Not to confident on Limits but I can usually get them, I just don't know how to get all the way there:

Let:

$$H = \infty$$

$$\frac {H}{\sqrt{4H^{2}+1}}$$

I don't know what to do from here to actually get a numerical result, but to me it looks like it would basically be H/sqrt(H^2) and should be one or zero but the actual answer they are looking for is 1/2 so could someone please instruct me on how to get there?

thanks!

You can never substitute infinity into a limit expression! Even when you cleverly hide it with a variable.

Factor t out of the numerator and denominator, and then take the limit.

 

[Asphyxiated]

Well I have been using the book "Elementary Calculus: An Infinitesimal Approach" and you do exactly that for the methods taught in the book. For example

Find:

$$\left ^{lim} _{x \rightarrow \infty} \right \frac {3x^{4}+5x -2}{2x^{4}-6x^{3}+7}$$

the way that it is taught in the book to solve this limit is:

$$\frac {3H^{4}+5H-2}{2H^{4}+6H^{3}+7} = \frac {3+5H^{-3}-2H^{-4}} {2-6H^{-1}+7H^{-4}}$$

and then a portion of this you are probably not familiar with is standard parts, the standard parts to an expression is the real number/variables and the 'hyperreal numbers/variables' are not part of the standard parts of an expression. and H is a hyperreal number and thus it is not taken when taking standard parts so to solve this, using the method taught in the book you would take:

$$st(\frac {3H^{4}+5H-2}{2H^{4}+6H^{3}+7}) = \frac{3+0-0}{2-0+0} =\frac {3}{2}$$

I want to learn the "standard" way of doing it though, but how can I factor out t from the denominator when it is in a binomial inside a square root?

 

[physicsman2]

Well I have been using the book "Elementary Calculus: An Infinitesimal Approach" and you do exactly that for the methods taught in the book. For example

Find:

$$\left ^{lim} _{x \rightarrow \infty} \right \frac {3x^{4}+5x -2}{2x^{4}-6x^{3}+7}$$

the way that it is taught in the book to solve this limit is:

$$\frac {3H^{4}+5H-2}{2H^{4}+6H^{3}+7} = \frac {3+5H^{-3}-2H^{-4}} {2-6H^{-1}+7H^{-4}}$$

and then a portion of this you are probably not familiar with is standard parts, the standard parts to an expression is the real number/variables and the 'hyperreal numbers/variables' are not part of the standard parts of an expression. and H is a hyperreal number and thus it is not taken when taking standard parts so to solve this, using the method taught in the book you would take:

$$st(\frac {3H^{4}+5H-2}{2H^{4}+6H^{3}+7}) = \frac{3+0-0}{2-0+0} =\frac {3}{2}$$

I want to learn the "standard" way of doing it though, but how can I factor out t from the denominator when it is in a binomial inside a square root?

The way I see it, that is the "standard" way. Observing the degrees of the variables in both the numerator and denominator and applying the limit at infinity is usually the way to go, like you showed in your example.

Because you have a binomial in the square root is the reason you can factor it out and take the square root of it to get t.

Do you understand?

 

[Asphyxiated]

ok well i think that you mean this:

$$\sqrt{4t^{2}+1} = \sqrt{(2t+1)(2t-1)}$$

but that works out to 4t^2-1 and (2t+1)(2t+1) is a trinomial, so... yeah

 

[physicsman2]

ok well i think that you mean this:

$$\sqrt{4t^{2}+1} = \sqrt{(2t+1)(2t-1)}$$

but that works out to 4t^2-1 and (2t+1)(2t+1) is a trinomial, so... yeah

Don't factor it, factor out a t^2. You should get sqrt(t^2(1 + 1/t^2)), right? I think you can go from there.

 

[Asphyxiated]

well I might have went around this a round-about way but basically it is like this:

$$\frac {t} {\sqrt{4t^{2}+1}}$$

$$\frac{t} {\sqrt{4t^{2}(1+\frac{1}{4t^{2}})}}$$

$$\frac{t} {(2t)\sqrt{1+\frac{1}{4t^{2}}}}$$

$$\frac{1} {(2)(\sqrt{1+\frac{1}{4t^{2}}}}$$

$$st(\frac{1} {(2)(\sqrt{1+\frac{1}{4t^{2}}})})$$

$$\frac {1}{2}$$

is that right?

 

[physicsman2]

well I might have went around this a round-about way but basically it is like this:

$$\frac {t} {\sqrt{4t^{2}+1}}$$

$$\frac{t} {\sqrt{4t^{2}(1+\frac{1}{4t^{2}})}}$$

$$\frac{t} {(2t)\sqrt{1+\frac{1}{4t^{2}}}}$$

$$\frac{1} {(2)(\sqrt{1+\frac{1}{4t^{2}}}}$$

$$st(\frac{1} {(2)(\sqrt{1+\frac{1}{4t^{2}}})})$$

$$\frac {1}{2}$$

is that right?

That is right. So it looks like you understand how to do it.

 

[Mark44]

Let's get the limit back in.
$$\lim_{t \to \infty} \frac{t}{\sqrt{4t^2 + 1}}= \lim_{t \to \infty} \frac{t}{t \sqrt{4 + 1/t^2}}= \lim_{t \to \infty} \frac{1}{1 \sqrt{4 + 1/t^2}} = \frac{1}{2}$$

For any finite value of t, t/t = 1, and this is also true in the limit. As t grows large without bound, the 1/t^2 term in the radical approaches zero, so the whole fraction approaches 1/sqrt(4) = 1/2.