limit at small values of n

[Anonymous No.11781429]

i understand it intuitively but is there some theorem or proof that can help me understand this better?

[Anonymous No.11781449]

>>11781429
the limit is 0

[Anonymous No.11781450]

>>11781449
What happens when a is 0

[Anonymous No.11781454]

>>11781450
then the limit is 1

[Anonymous No.11781456]

>>11781450
l'hospital's rule

[Anonymous No.11781459]

>>11781456
lol nope

[Anonymous No.11781464]

>>11781459
I will solve that limit with l'hospital's rule all I damn well want and there's nothing you can do to stop me.

[Anonymous No.11781480]

>>11781429
N/(a+N)=(a+N-a)/(a+N)=1 - a/(a+N)
Replace n with zero, now clearly a/a is 1 and you are just dealing witth finite quantities so 1-1 is 0.

[Anonymous No.11781498]

>>11781449
yes, but n is 0, so technically the formula is n/a

[Anonymous No.11781500]

>>11781498
if n = 0, then **technically** n/(n+a) = 0/(0+a) = 0. why would you plug in n = 0 only for one instance of n ?

[Anonymous No.11781522]

>>11781450
[math]
\displaystyle
\lim_{n \to 0} \dfrac{n}{a+n}
= \lim_{n \to 0} \dfrac{ \frac{d}{dn}n}{ \frac{d}{dn}(a+n)}
= \lim_{n \to 0} \dfrac{1}{0+1}=1
[/math]

[Anonymous No.11781549]

>>11781429
Just show that the function x/(a + x) is continuous at x = 0.

If you do that, then the limit as x approaches 0 will be the same as just plugging 0 in for x.

However, if x/(a + x) is not continuous at x = 0 (this happens when a = 0), then you can't just plug x = 0 in, and you will have to come up with a different strategy for figuring out the limit.

[Anonymous No.11781552]

>>11781522
L'hospital's rule only works when the fraction is indeterminate, i.e., when its 0/0, or infinity/infinity.
This is wrong when a =/= 0.

[Anonymous No.11781554]

>>11781500
calculus is meant to simplify things that would otherwise be atrocious in algebra.

[Anonymous No.11781566]

>>11781429
Take a look in this pdf
https://people.kth.se/~tek/envarre99/teori1.pdf

[Anonymous No.11781573]

>>11781522
good, because I fucking answered >>11781450

[Anonymous No.11781575]

>>11781552
good, because I fucking answered >>11781450

[Anonymous No.11781620]

>>11781554
no idea what you're trying to say

[Anonymous No.11781706]

>>11781429
The limit of a division is the division of limits.
lim n as n-> 0 is 0
lim a+n as n->0 is a
therefore lim n/(a+n) as n-> 0 is: 0/a which is just 0

[Anonymous No.11781802]

>>11781620
obviously i'm just trolling bud, relax.

[Anonymous No.11781854]

>>11781429
Taylor expand around n=0?

[Anonymous No.11781889]

>>11781429
[math]n/a[/math] is an approximation for [math]n << a[/math]. The limit as n approaches 0 is 0 for nonzero a and nonexistent for [math]a = 0[/math]

[Anonymous No.11783367]

>>11781450
[math]\lim_{n \to 0} ({1-n})=0,9999999999...=1[/math]

[Anonymous No.11783413]

OP here, I just said that since for small values of N, N+A = A, "lim_n->inf N/N+A = N/A. truly in physicist fashion

[Anonymous No.11783441]

>>11783413
Nope, some quotients of infinity are undefined.

[Anonymous No.11784381]

>>11781464
>not using l'ambulance rule

[Anonymous No.11784383]

>>11784381
its L'Hôpital's rule

[Anonymous No.11784393]

>>11784383
can't use it bro they're all full of covid patients

[Anonymous No.11784543]

>>11781450
n/n = 1 obviously

[Anonymous No.11784559]

>>11781522
you don't know that a isn't a function of n though, so the derivative might be nonzero.

[Anonymous No.11784764]

>>11783413
For non-zero values of a, they both tend to 0 and so they tend to each other. But that’s not particularly interesting, right? I mean, you could say the same thing about n^2/a or even just n.

I think the point here is a bit more profound which is that the ratio of the two quantities goes to 1. Depending on the context, it means that one is a good approximation for the other in a numerical sense that’s different from either of my two examples.

[Anonymous No.11785030]

>>11781522
To bad n is approaching 0 and not 1

[Anonymous No.11785041]

>>11785030
lol
lrn2math

[Anonymous No.11785045]

>>11784559
lrn2read >>11781450

[Anonymous No.11785679]

this is unrelated anons but could you please help me solve this world problem using the polyas process> idk how to solve it yes im retarded in math:
Four children, Amy, Susie, Tessie, and Eddie are lined up according to height, each holding a balloon. The child in front (the shortest) is holding neither a red nor blue balloon. Susie is holding a red balloon. Tessie sees exactly two balloons in front of her. The child holding the blue balloon is right in front of the child with the yellow balloon. Amy is in front of Tessie. One child is holding a white balloon. Determine the arrangement of the four children (from shortest to tallest) and the color of the balloon they are holding.

I already tried reddit but no one answered it. might have been too easy for them idk. please help me...

[None !!IlL8ybF38OS No.11787272]

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