1. ## Find a limit

Hello every one,

I need help to understand how this limit is evaluated:

lim.jpg
I found it in a physics text book but I could not understand how they did it.

Thanks

2. Originally Posted by Viona
Hello every one,

I need help to understand how this limit is evaluated:

lim.jpg
I found it in a physics text book but I could not understand how they did it.

Thanks
First, please read this summary of what we do, how we do it, and what you must do to get help.

https://www.freemathhelp.com/forum/t...l=1#post433156

In this case, we particularly need to know whether you have studied limits in a calculus or a pre-calculus course.

Second, we need more context because what you have shown us is odd. Perhaps there is more that you have not shown.

Third, please tell us what you have tried or thought while trying to understand.

EDIT: What you have been shown is not math, but intuition.

If x is very large, then

$\dfrac{2(x + a + 1) - b}{x} = \dfrac{2x}{x} + \dfrac{2a + 2 - b}{x} \approx 2 + 0 = 2.$

Similarly, if x is very large,

$\dfrac{x + 2a + 2}{x} = \dfrac{x}{x} + \dfrac{2a + 2}{x} \approx 1 + 0 = 1.$

So if x is very large

$\dfrac{2(x + a + 1) - b}{(x + 1)(x + 2a + 2} = \dfrac{\dfrac{2(x + a + 1) - b}{x}}{ (x + 1) * \dfrac{x + 2a + 2}{x}} \approx \dfrac{2}{(x + 1) * 1} = \dfrac{2}{x + 1}.$

This is kind of foolish however because if x is very large, the resulting approximation is itself is approximately zero.

3. Originally Posted by Viona
Hello every one,

I need help to understand how this limit is evaluated:

lim.jpg
I found it in a physics text book but I could not understand how they did it.

Thanks
Clearly this is not actually a limit, since x can't be in the result. It is something more like an asymptotic approximation.

4. Originally Posted by JeffM
First, please read this summary of what we do, how we do it, and what you must do to get help.

https://www.freemathhelp.com/forum/t...l=1#post433156

In this case, we particularly need to know whether you have studied limits in a calculus or a pre-calculus course.

Second, we need more context because what you have shown us is odd. Perhaps there is more that you have not shown.

Third, please tell us what you have tried or thought while trying to understand.

EDIT: What you have been shown is not math, but intuition.

If x is very large, then

$\dfrac{2(x + a + 1) - b}{x} = \dfrac{2x}{x} + \dfrac{2a + 2 - b}{x} \approx 2 + 0 = 2.$

Similarly, if x is very large,

$\dfrac{x + 2a + 2}{x} = \dfrac{x}{x} + \dfrac{2a + 2}{x} \approx 1 + 0 = 1.$

So if x is very large

$\dfrac{2(x + a + 1) - b}{(x + 1)(x + 2a + 2} = \dfrac{\dfrac{2(x + a + 1) - b}{x}}{ (x + 1) * \dfrac{x + 2a + 2}{x}} \approx \dfrac{2}{(x + 1) * 1} = \dfrac{2}{x + 1}.$

This is kind of foolish however because if x is very large, the resulting approximation is itself is approximately zero.

Yes I studied limits long time ago, but this one seemed peculiar to me. Now I think it is some kind of asymptotic behavior. Thank you

5. Originally Posted by Dr.Peterson
Clearly this is not actually a limit, since x can't be in the result. It is something more like an asymptotic approximation.

I think you are write. I misunderstood what they mean. This actually what the book said:
rec.jpgThanks.

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