What Is the Limit of t Plus a Complex Fraction as t Approaches Infinity?

[Aleolomorfo]

Homework Statement

Finding the value of the limit:
$$\lim_{t\to +\infty} t+\frac{1-\sqrt{1+a^2t^2}}{a}$$
$a$ is just a costant

The Attempt at a Solution

At first sight I had thought that the limit was $\infty$ but then I realized that there is an indeterminate form $\infty - \infty$. I tried to rationalize the second part of the sum in the limit and I found $\frac{\infty}{\infty}$ which gave $\infty$ and so I found again $\infty - \infty$. I don't really know how to get the answer.

 

[StoneTemplePython]

There are a few approaches. I like to do qualitative stuff first -- i.e. first demonstrate that this converges (specifically that it has an upper bound in this case).

You understand that
$\sqrt{1+a^2 t^2} \geq \sqrt{a^2 t^2} = at$

right? So if instead of subtracting the bigger number we subtract the smaller one, we get a larger result, giving us

$t+\frac{1-\sqrt{1+a^2t^2}}{a} \leq t+\frac{1-\sqrt{a^2t^2}}{a} = ?$
simplify that and you should get a nice upper bound that does not involve $t$. The challenge is then to do the quantitative part and show that you actually reach this upper bound.
- - -
edit: I should have pointed out I was assuming $a \gt 0$.

If $a$ is negative we have a problem. I.e. in the event of $a \lt 0$

$t+\frac{1-\sqrt{1+a^2t^2}}{a} = \frac{1}{a}+t-\frac{\sqrt{1+a^2t^2}}{a} = \frac{-1}{\big \vert a\big \vert}+t +\frac{\sqrt{1+a^2t^2}}{\big \vert a\big \vert}$

which grows to infinity as t grows large.

 

[Ray Vickson]

Homework Statement

Finding the value of the limit:
$$\lim_{t\to +\infty} t+\frac{1-\sqrt{1+a^2t^2}}{a}$$
$a$ is just a costant

The Attempt at a Solution

At first sight I had thought that the limit was $\infty$ but then I realized that there is an indeterminate form $\infty - \infty$. I tried to rationalize the second part of the sum in the limit and I found $\frac{\infty}{\infty}$ which gave $\infty$ and so I found again $\infty - \infty$. I don't really know how to get the answer.

Assume that $a >0$ (the argument with $a < 0$ is similar) and note that we are interested in positive values of $t$.

Use the fact that
$$\sqrt{1+a^2 t^2} = \sqrt{a^2 t^2} \sqrt{ 1 + \frac{1}{a^2 t^2} } = a t \sqrt{ 1 + \frac{1}{a^2 t^2} }$$
Since $1/(a^2 t^2) \ll 1$, the last square-root is almost 1. However, we need more: we need to know how the last square-root differs from 1 for large, but finite values of $t > 0$. Are you familiar with the result that $\sqrt{1+x} = 1 + (1/2) x + r(x)$, where $|r(x)/x| \to 0$ as $x \to 0$? Apply that, with $x = 1/(at)^2$.

 

[PeroK]

Homework Statement

Finding the value of the limit:
$$\lim_{t\to +\infty} t+\frac{1-\sqrt{1+a^2t^2}}{a}$$
$a$ is just a costant

The Attempt at a Solution

At first sight I had thought that the limit was $\infty$ but then I realized that there is an indeterminate form $\infty - \infty$. I tried to rationalize the second part of the sum in the limit and I found $\frac{\infty}{\infty}$ which gave $\infty$ and so I found again $\infty - \infty$. I don't really know how to get the answer.

Assuming you are not doing Real Analysis and you just want to calculate the limit, rather than prove it rigorously. (I guess it came from your SR acceleration problem.). Then you could just do the following:

For large $t$, $\sqrt{1+a^2t^2} \approx at$.

 

[Aleolomorfo]

First of all, thank you very much for your help, all of you.

There are a few approaches. I like to do qualitative stuff first -- i.e. first demonstrate that this converges (specifically that it has an upper bound in this case).

You understand that
√1+a2t2≥√a2t2=at1+a2t2≥a2t2=at\sqrt{1+a^2 t^2} \geq \sqrt{a^2 t^2} = at

right? So if instead of subtracting the bigger number we subtract the smaller one, we get a larger result, giving us

t+1−√1+a2t2a≤t+1−√a2t2a=?t+1−1+a2t2a≤t+1−a2t2a=?t+\frac{1-\sqrt{1+a^2t^2}}{a} \leq t+\frac{1-\sqrt{a^2t^2}}{a} = ?
simplify that and you should get a nice upper bound that does not involve ttt. The challenge is then to do the quantitative part and show that you actually reach this upper bound.
- - -
edit: I should have pointed out I was assuming a>0a>0a \gt 0.

If aaa is negative we have a problem. I.e. in the event of a<0a<0a \lt 0

t+1−√1+a2t2a=1a+t−√1+a2t2a=−1∣∣a∣∣+t+√1+a2t2∣∣a∣∣t+1−1+a2t2a=1a+t−1+a2t2a=−1|a|+t+1+a2t2|a|t+\frac{1-\sqrt{1+a^2t^2}}{a} = \frac{1}{a}+t-\frac{\sqrt{1+a^2t^2}}{a} = \frac{-1}{\big \vert a\big \vert}+t +\frac{\sqrt{1+a^2t^2}}{\big \vert a\big \vert}

which grows to infinity as t grows large.

It is a bit too mathematical way to approach the problem from my physical perspective, but it is very interesting. Mathematicians always fascinate me.

Assume that a>0a>0a >0 (the argument with a<0a<0a < 0 is similar) and note that we are interested in positive values of ttt.

Use the fact that

√1+a2t2=√a2t2√1+1a2t2=at√1+1a2t21+a2t2=a2t21+1a2t2=at1+1a2t2​

\sqrt{1+a^2 t^2} = \sqrt{a^2 t^2} \sqrt{ 1 + \frac{1}{a^2 t^2} } = a t \sqrt{ 1 + \frac{1}{a^2 t^2} }
Since 1/(a2t2)≪11/(a2t2)≪11/(a^2 t^2) \ll 1, the last square-root is almost 1. However, we need more: we need to know how the last square-root differs from 1 for large, but finite values of t>0t>0t > 0. Are you familiar with the result that √1+x=1+(1/2)x+r(x)1+x=1+(1/2)x+r(x)\sqrt{1+x} = 1 + (1/2) x + r(x), where |r(x)/x|→0|r(x)/x|→0|r(x)/x| \to 0 as x→0x→0x \to 0? Apply that, with x=1/(at)2x=1/(at)2x = 1/(at)^2.

Your advice is very helpful, sometimes I forget the simpliest method to do mathematical exercises. The result I have found is $1/a$ and I think it is correct, but I wait your response.

Assuming you are not doing Real Analysis and you just want to calculate the limit, rather than prove it rigorously. (I guess it came from your SR acceleration problem.). Then you could just do the following:

For large ttt, √1+a2t2≈at1+a2t2≈at\sqrt{1+a^2t^2} \approx at.

This was my first approach but it was not very successfull.

 

[SammyS]

First of all, thank you very much for your help, all of you.
...

Your advice is very helpful, sometimes I forget the simplest method to do mathematical exercises. The result I have found is $1/a$ and I think it is correct, but I wait your response.

1/a is also what I get.

Added in Edit:
As @StoneTemplePython mentioned in Post #2, you need a > 0 for this limit to be finite.

 

[Gene Naden]

I agree with your answer, if you take the positive square root. If you take the negative square root then the t's don't cancel and that is a problem.

 

[haruspex]

I agree with your answer, if you take the positive square root. If you take the negative square root then the t's don't cancel and that is a problem.

By definition, the function √ returns the positive square root.

 

[Mark44]

By definition, the function √ returns the positive square root.

A discussion we've had many times over the years here at this site.

 

[Gene Naden]

I stand corrected

 

[Ray Vickson]

I agree with your answer, if you take the positive square root. If you take the negative square root then the t's don't cancel and that is a problem.

In general, for large $t > 0$ we have $\sqrt{1 + a^2 t^2} \sim |a| t.$