mathscribbles
Region: US
Tuesday 23 September 2025 20:58:35 GMT
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k_nel24 :
first one instantly 0 no?
2025-09-23 21:02:57
45
xskr1pt :
bounded * 0 = 0, squeeze theorem
2025-09-24 12:22:44
14
Pakistani living in 🇿🇦 :
All of that work just to get zero
2025-09-24 23:07:45
7
🦕u :
can you use squeeze theorem also instead of the algebraic manipulation?
2025-09-23 21:15:21
1
Torabora77 :
I don’t like this! This first step doesn’t make sense to me: lim sin(1/x) / (1/x) is not 1 when x approaches zero. It is true if x goes to ♾️or (1/x) goes to zero!!
2025-09-24 15:19:10
6
Jay don’t play :
Just use the squeeze theorem it makes it so much easier
2025-09-30 03:30:06
2
maayan :
bounded*0=0
2025-09-23 22:11:24
5
gech :
the first is incorrect there is no infinitesimal inside the sin and cos so you cannot use famous limits like sinx/x or 1-cosx/x but you have to use taylor to infinity so only knowing that sin1/x and cos1/x are between 1 an -1 something that is not infinity multiples something that is almost zero is zero
2025-09-24 14:53:11
4
ksmith :
0 / 1/x is x 🤔
2025-10-21 04:40:04
0
tomerr :
you used the sine limit wrongly as you have 1/x and not x, the limit is still 0
2025-09-25 07:19:07
1
PinkyTrent :
Would it be allowed to use exponential identity for both sinx and cosx and approach from there?
2025-10-01 08:23:34
0
god_of_the_star :
squeeze theorem
2025-10-04 10:06:23
0
smburns47 :
both sin and cos are bounded by +-1 so x^2 times a constant goes to zero as x goes to zero
2025-09-24 15:41:56
0
Райтон Иячоващо :
by graphing it on a calculator, i can see that it reaches 0 because they are trying to be multiplied by a term like x². meaning that they depend on x² unless x²≠0.
2025-09-28 18:45:16
0
Жомарт Онербек :
Thanks, very interesting!!
2025-10-02 01:44:47
0
MICHAEL ✪ :
Integration (X^4)((1-root x)^5) from 0 to 1 with respect to X
2025-09-24 07:50:52
0
ricardorib :
what happen to the - in the cos.
2025-10-17 20:24:41
0
Simon Zilouf :
😁😁😁
2025-10-21 14:20:11
0
BlueBoy22 :
🥀
2025-10-07 12:49:55
0
Felix Svensson :
🥰🥰🥰
2025-12-19 04:16:26
0
nicole :
why is sin(1/x)/(1/x) = 1 as x -> 0? doesnt the 1/x go to infinity, which is diff from the sinx/x result?
2025-09-23 21:42:06
6
@krakrak🏳️🌈🇵🇱🇪🇺 :
When f is bounded, then lim x*f(x) =0 for x->0, even more so for x². Sin+cos is bounded
2025-09-23 21:49:50
0
43567kkpp :
:direct substituion will work for this case since the domain of sin and cos is " Real number" so it wouldnt matter even if denominatior is 0 so it would be 0 (sin (1/0) + cos(1/0) ) which is 0.
2025-09-23 21:28:20
2
3moor :
Wait this might be a dumb question but isn’t this just zero? If you plug in zero, you get 0^2, which makes the whole expression zero regardless of what’s inside the brackets. So the result is simply zero.
2025-09-24 13:15:01
0
@krakrak🏳️🌈🇵🇱🇪🇺 :
While the end result is correct here. You cant just take part of the formula, and replace it with the the limit, while still calculating the limit. For example, with x->0,the limit of x*(1/x) is 1. You cant say that the limit of x is 0,so it's lim 0*(1/x)=0
2025-09-23 21:54:54
0
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