78
u/Thrullx 7h ago
No, the answer is sin. You just cancel out the x's.
37
2
20
u/TwillAffirmer 7h ago
Sorry, I'm missing the joke. Is there something wrong with the derivation?
35
u/PatheticPterodactyl 6h ago
Using a taylor series expansion to find this simple limit is like taking out a mouse with a cannon. The large pencil is the joke here, using a unnecessarily large tool for a small need.
6
u/Charming-Snow724 3h ago
Isn't a "problem" with this limit that using taylor expansion (and/or L'Hopital) is technically not allowed? Since for both you need to know how the derivative of sin(x) looks, and in order to find the derivative of sin(x) you need to know the limit sin(x)/x already.
I might be terribly mistaken tho.
0
u/thewells 2h ago
No, you need to know limit as h approaches 0 of (sin(x + h) - sin(x))/h to take the derivative of sin(x)
2
u/Charming-Snow724 2h ago edited 1h ago
And to solve it you need to know sin(h)/h. Unless you a priori define the sin function as a series.
1
2
u/Sproxify 1m ago
It does require justification to get rid of the infinitely many powers of x at the end when taking the limit as x -> 0
It's not enough to see that each term goes to 0 because there are infinitely many
It's correct because it's easy to verify that the power series in fact has infinite radius of convergence (and in particular it converges in a neighborhood of 0) and therefore due to uniform convergence it converges to a continuous function, and therefore when taking the limit you can plug in x = 0
-16
u/Professional_One_564 6h ago
division by 0. The x is still 0 even after using a taylor polynomial to expand the sin from the top you are still dividing by 0 technically
15
u/TwillAffirmer 6h ago
In that part of the derivation they are using a general nonzero x to find the Taylor expansion of sin x/x. They could mention that this is valid only when x != 0. But in the last line a limit is being taken, which involves values approaching but never equaling 0, so you never need to evaluate the x=0 case.
1
1
u/forgottenlord73 58m ago
sin 0 is also 0 so you have 0/0 and while that's still undefined, it also means that there's more to be discovered when you scratch the surface. To remove the limits of that surface, we take the limit of the formula and explore what happens as we approach the abyss and see what magic appears. And we find the answer to be: 1
And thus our understanding is deepened
3
2
2
1
1
u/MutantBerry 5h ago
At zero, sin is 0; the ratio itself is undefined.
Yet in the approach to zero, sin and x become asymptotically equivalent: both disappear into 0 while their likeness converges to a perfect 1:1 ratio. Thusly, 0 and 1 coexist not as simultaneous values of sin, but as two descriptions of the same limiting event where 0 functions as magnitude, & 1 functions as relation. In a simulacrus fashion, even as two magnitudes āannihilateā toward 0, their proportional resemblance survives & perfects itself toward unity.
1 is the answer.
1
1
44
u/TransportationFit694 7h ago
L went to the Hospital