Çözüldü Trigonometric Limit Without Using L'Hospital's Rule - Trigonometric Conversions - Taylor Series

lim (x → π / 2) [ 1 - sin(x) ] / (2x - π) = ?

https://i.servimg.com/u/f72/19/97/10/39/limit13.png
https://www.facebook.com/photo?fbid=260873822121773&set=gm.2151979901606208
(One of the teachers in that group suggested an incorrect and inapplicable usage of series.)
(That Facebook group was changed to private on September 25, 2022, so only the members can see if there is any other solution.)

0 / 0 indetermination

Solution - 1
lim (x → π / 2) [ sin(π / 2) - sin(x) ] / [ -(π - 2x) ] =
-lim (x → π / 2) 2·{ sin[ (π / 2 - x) / 2 ]·{ cos[ (π / 2 + x) / 2 ] } / (π - 2x) =
-lim (x → π / 2) 2·{ sin[ (π - 2x) / 4 ] }·{ cos[ (π + 2x) / 4 ] } / [ 4·(π - 2x) / 4 ] =
-(1 / 2)·lim (x → π / 2) { sin[ (π - 2x) / 4 ] / [ (π - 2x) / 4 ] } · lim (x → π / 2) cos[ (π + 2x) / 4 ] =
Using (π - 2x) / 4 = y variable conversion; x → π / 2 ⇒ y → 0
-(1 / 2)·[ lim (y → 0) sin(y) / y ]·cos[ (π + π) / 4 ] =
-(1 / 2)·1·cos(π / 2) =
-(1 / 2)·1·0 =
0
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Solution - 2
x - π / 2 = p
lim (p → 0) [ 1 - sin(π / 2 + p) ] / (2p) =
lim (p → 0) [ 1 - cos(p) ] / (2p) =
lim (p → 0) { 1 - 1 + 2·[ sin(p / 2) ]^2 } / (2p) =
lim (p → 0) { [ sin(p / 2) ]^2 } / p =
lim (p → 0) { [ sin(p / 2) ]^2 } / [ (p^2 / 4)·4p ] =
lim (p → 0) { [ sin(p / 2) ] / (p / 2) }^2 · lim (p → 0) 4p =
(1^2)·4·0 =
0
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Solution - 3
For y = f(x) = [ 1 - sin(x) ] / (2x - π), using Taylor expansion at x = π / 2
(this is homework for the interested students, if there are any left around, to study, understand, and apply the information given at https://tutorial.math.lamar.edu/classes/calcii/taylorseries.aspx);

f(x) = (1 / 4)·(x - π / 2) - (1 / 48)·(x - π / 2)^3 + (1 / 1440)·(x - π / 2)^5 + ...
lim (x → π / 2) f(x) = (1 / 4)·0 - (1 / 48)·0^3 + (1 / 1440)·0^5 + ... =
0 - 0 + 0 - ... =
0
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Note: No need to be checked by WolframAlpha at all.