A Nice Limit Done Without Using Wolframalpha

[cblpu]

You had one of the best math channels on youtube for learning. Please come back.

[jakubwieliczko]

Hello. For the past 3 years I have been preparing for Polish Mathematical Olympiad. Ever since I started working on it, it was my dream to participate in the final round. You were one of the first youtubers i stumpled upon when doing research and I’ve watched you ever since. Last week I participated in the 2nd round of PMO and it turned out very well. I’m still not sure if I will qualify but it’s likely, given how I performed compared to other people. That’s why I wanted to thank you for all the good work you’ve done by running this channel and wherever you are right now, I hope everything is good for you.
Best wishes from your fan

[hitzcritz]

This is by far the greatest math channel on YouTube. I really miss you

[ВасяДенисов-хм]

Where are you? Please, come back. You are one of the best math channel on YouTube.

[johnchessant]

I'm still waiting to see you next time :)

[Mephisto]

4:34 u is infinity only if t is zero plus, otherwise it is negative infinity.

[billbill]

Why did u stop uploading???! Your content is amazing and we all want you back if possible. Keep up the good work!!

[bhavyachobisa]

Pls do comeback u are one the best math solving channels

[yoav]

Sin(1/(x-1)) is bounded ([-1, 1]) so you could just use lohpital for (x-1)^2/sin(pix) and you get 2(x-1)/(picos(pix)) which is 0/-pi =0 so the limit is 0.
You can put any thing you want in the sin(1/(x-1)) and still get 0 .

[karangupta]

Try solving this functional equation:

Find all f: R+ --> R+ such that
f(f(x)/y) = y•f(y)•f(fx))
for all x, y belonging to R+.

[Coileain-frs]

You are the best, come back and do it again

[marcoaltamura]

sin (pi x ) = sin (pi - pi x)

Then we have

lim x->1 ((x-1)²*sin(1/(x-1)))/sin(pi - pi x)

The denominator becomes simply
pi - pi x (approaching 1 this is equivalent to the original denominator). We can factor out a minus pi and get - pi (x-1), which can be simplyfied with the numerator to get:

lim x->1 ((x-1)*sin(1/(x-1)))/-pi)

In the numerator we have, by substituting, 0 times a sin function which oscillates between -1 and 1, so the product of those is still 0.

0 over a finite number like -pi is simply 0.


So the limit is equal to 0.

[stanleydodds]

Depends on the domain of the function. For real x, the limit exists (and is trivially 0) as sin is bounded for the numerator and has a simple order 1 pole for the denominator (i.e. Squeeze plus L'Hopital).

For complex x, the limit does not exist (for instance, it diverges if we approach 1 from the imaginary direction). Also, the pole at 1 is essential, because of the nasty function sin(1/(1-x)).

[MrMinime]

way too complicated with those backwards and forwards substitutions

1. *sin(πx) = sin(πx - π + π) = -sin(πx - π)=-sin(π(x-1))*
==> *(x-1)/sin(πx) = -(x-1)/sin(π(x-1) = -1/π * (π(x-1)/sin(π(x-1) --> -1/π* when x-->1

2. *(x-1)*sin(1/(x-1)) --> 0* (obviously, as sin is limited by [-1..1] and *(x-1)-->0* )

3. So we have a product of 2 factors where the 1st factor *--> -1/π* and the 2nd factor *-->0* ==> The total limes is obviously 0

(and strictly speaking the deduction "lim(a/b) = lim(a)/lim(b)" at 4:09 is in general wrong and must be justified)

[calebtolman]

lim f(x) as x -> 0+ is the same as lim f(1/x) as (1/x) -> infinity. If you are taking lim f(x) as x -> 0, then you need to check both lim f(1/x) as x -> +infinity and also check lim f(1/x) as (1/x) -> -infinity and the values must match. In this case they are both 0 so the limit exists but in general that doesn't always happen

[achrafbaiz]

There is way lot easy than this
But in general your videos are greats

[ItsJustEthan]

4:23 ish I think you have to consider the fact that the lim as t approaches 0 u approaches + OR MINUS infinity. Don't think this matters because that limit would still approach 0 after evaluating the other case so I think the result is still 0 but I think you need this to be rigorous.

[advaypakhale]

What software do you use to create these videos? It's great!

[mdperpe]

After the substitution t = x-1 we get the expression t² sin(1/t) / sin(π(t+1)) which can be rewritten as -(t/π) (πt/sin(π(t+1))) sin(1/t).
Now, -|(t/π) (πt/sin(π(t+1)))| ≤ -(t/π) (πt/sin(π(t+1))) sin(1/t) ≤ +|(t/π) (πt/sin(π(t+1)))| where both the first and last expressions have limits 0*1 = 0.
Therefore, by the squeeze theorem, also -(t/π) (πt/sin(π(t+1))) sin(1/t) has limit 0.

Can we just use bounded [-1, 1] for sin (1/(x-1)) and remove it?