#Derivatives and values
opal windBOT
lament geyser
Try using limit definitions
hmm if my work is correct then f ends up being a trivial function for condition 3 to be the case 
eager frost
hmm if my work is correct then f ends up being a trivial function for condition ...
I tried that yeah, but could you maybe explain a bit more?
I’ve been using limit -> 0 and squeeze theorem, but still can’t quite figure it out
vast briar
For question 2)
We can first start by attempting to evaluate (f(0+h)-f(0))/h and try to see if this limit exists as h goes to zero
(h^2 sin(A/h)-B)/h
= h sin(A/h) - B/h
eager frost
(h^2 sin(A/h)-B)/h
= h sin(A/h) - B/h
Wouldn’t that end up being 0?
eager frost
Oh wait, the sin diverges
vast briar
Unless...
eager frost
Yeah, I should know this, but I don’t lol…
vast briar
Does B/h always diverge ?
eager frost
Doesn’t have to, I guess as we need to find an answer
vast briar
Does 0/h diverge ?
eager frost
Yes
eager frost
Does 0/h diverge ?
I mean it would always be 0 or undefined right, 1/x diverges as well
eager frost
Does 0/h diverge ?
Or am I just totally clueless rn?
eager frost
Why would it be undefined
I mean 1/x diverges as well, so I therefore thought that 0/x would too
lament geyser
0/h doesn't diverge, for limits you only look at the punctured neighborhood around a
I. E. In this instance assume 0 <|h|
So the limit is just 0
eager frost
Wait, so both values are 0?
eager frost
So the limit is just 0
Because this means both limits go to 0, don’t A and B then have to be 0?
lament geyser
For part 3 yeah p much 
part 2 I think A can be the value you found for A in part 1. Which might also be 0 if B is 0 i haven't checked
eager frost
In part 1 I found that A can be any real number, meaning that will also apply to this case then. B = 0, judging from the limits
eager frost
So A is any real number and B = 0?
lament geyser
mhm
eager frost
For part 3 yeah p much <:devastation:898725325335298088>
Cool, but why did you say that for 3 both are equal to 0?
lament geyser
uhh well when you take the derivative around x \neq 0, you'll see that the only way the limit as x goes to 0 can even exist is if A = 0. If I did the math right, which I might've not 🤷♂️
eager frost
I see, well I’ll first make sure I get part 2 done correctly
So, because the limit of the first term is 0 and the second too, A can be any real number as that doesn’t matter, while B end up being 0 as well
lament geyser
mhm
eager frost
Lol, the same solution 3 times in a row just can’t be right
I think for 3 A can be any real number as well and B = 0
lament geyser
hm? When I derived I get $2xsin(A/x) + (-1)Ax^2/x^2cos(A/x)= 2xsin(A/x) - Acos(A/x)
Then the limit for continuity can only exist if A =0 cus of the cos(A/x) term
eager frost
When I tried the derivative I got 2xsin(A/X) -Acos(A/x)
eager frost
hm? When I derived I get $2xsin(A/x) + (-1)Ax^2/x^2cos(A/x)= 2xsin(A/x) - Acos(A...
Maybe something’s wrong with my derivative though
lament geyser
When I tried the derivative I got 2xsin(A/X) -Acos(A/x)
Oops I forgot to include the A but we got the same thing
So yeh cos(A/x) is gonna oscillate at a higher and higher frequency as x goes to 0, i. E. There won't be a limit unless A = 0
eager frost
Okay, so the derivative is 2xsin(A/x) - Acos(A/x) right?
eager frost
So yeh cos(A/x) is gonna oscillate at a higher and higher frequency as x goes to...
The limit of the first term = 0, and for the second A = 0, because otherwise there won’t be a limit?
lament geyser
Ye
eager frost
How should I phrase that properly lol?
Because I understand in my head what you’re saying, but can’t quite put it into word
lament geyser
well you could write out what f'(x) is first
lament geyser
But to prove continuity can’t I also just use left and right limit?
left and right limits don't exist either unless A = 0,still have infinite oscillation going on
Well I guess that's the part that's hard to write precisely
eager frost
Yeah I can write that lim x -> 0 of 2xsin(a/x) = 0, but not sure about the second term
lament geyser
I guess you could say let (a_n = A/(pi*n)) for A \neq 0, which tends towards 0. If f'(x) were continuous, then the limit of cos(A/a_n) = cos(pi*n) would exist as n goes to infinity. However, cos(pi*n) oscillates between 1 and - 1 as n goes to infinity
eager frost
But how would that imply A = 0?
eager frost
I guess you could say let (a_n = A/(pi\*n)) for A \neq 0, which tends towards 0....
Like can’t I just write something without that sequence? As in to define a limit for this function A = 0? Otherwise it doesn’t exist because of oscillation
lament geyser
But how would that imply A = 0?
A = 0 reduces it to cos(0)
eager frost
A = 0 reduces it to cos(0)
Which is 1, but isn’t that what you said before? Like A needs to be 0 such that a limit can exist
lament geyser
yeah ur also multiplying by A = 0 so that makes the whole limit equal 0 as x->0, hence continuous if B=0
eager frost
yeah ur also multiplying by A = 0 so that makes the whole limit equal 0 as x->0,...
Yeah so can’t I write it out like that or do you have a different way to go about it?
lament geyser
idk I was just proving why it oscillates
if u don't have to justify that then ur good
eager frost
if u don't have to justify that then ur good
Yeah I don’t have to prove why it oscillates, so then I can just write A = 0?
lament geyser
sure
eager frost
sure
Cool, thank you so much for your help! I finally solved them, thanks to your help
lament geyser
mhm
eager frost
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