Find a limit

Viona

New member
Joined
Nov 22, 2018
Messages
21
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
 

JeffM

Elite Member
Joined
Sep 14, 2012
Messages
3,258
Hello every one,

I need help to understand how this limit is evaluated:

View attachment 10614
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/threads/112086-Guidelines-Summary?p=433156&viewfull=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

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

Similarly, if x is very large,

\(\displaystyle \dfrac{x + 2a + 2}{x} = \dfrac{x}{x} + \dfrac{2a + 2}{x} \approx 1 + 0 = 1.\)

So if x is very large

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

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

Dr.Peterson

Elite Member
Joined
Nov 12, 2017
Messages
3,361
Hello every one,

I need help to understand how this limit is evaluated:

View attachment 10614
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.

Please quote what was said leading up to, or explaining, this.
 

Viona

New member
Joined
Nov 22, 2018
Messages
21
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/threads/112086-Guidelines-Summary?p=433156&viewfull=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

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

Similarly, if x is very large,

\(\displaystyle \dfrac{x + 2a + 2}{x} = \dfrac{x}{x} + \dfrac{2a + 2}{x} \approx 1 + 0 = 1.\)

So if x is very large

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

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
 

Viona

New member
Joined
Nov 22, 2018
Messages
21
Clearly this is not actually a limit, since x can't be in the result. It is something more like an asymptotic approximation.

Please quote what was said leading up to, or explaining, this.
I think you are write. I misunderstood what they mean. This actually what the book said:
rec.jpgThanks.
 
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