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Calculus/sqeeze theorem
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Question
how can i prove this squeeze theorem?
lim sin t
x approaches 0 ______=1
t
we will prove it in 3 different ways:
1) Lopital's low
Lim [f(x)/g(x)] = Lim [f’(x)/g’(x)]
Lim sin(t)/t= Lim cos(t)/1 = 1 when t = 0
2) Tlaylor series
The function sin(t) can be written as
t- t^3/6+t^5/120+....
So sin(t)/t =1-t^2/6+t^4/120+...
Lim (t-->0) = 1+0+0+…=1
3) For t very small, the form sin(t)< t< tang(t)
Is very true. The form can be written as
Sin(t)<t & t< sin(t)/cos(t).
Sin (t)< t --> sin(t) /t<1
t<sin(t)/cos(t)-->sin(t)/t>cos(t)
So now we have the law of ‘sandwich’:
Lim cost = 1 & Lim 1=1 then Lim sin(t)/t = 1
Remarks :
1. The form sin(t)< t< tang(t) can be proved by drawing a unit circle where x is the angle.
2. ‘Sandwich’ rule : If g(x)<f(x)<h(x) & Lim g(x)=Lim h(x)=L then Lim f(x)=L
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Answers by Expert:
Alon Mandes
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