It is easy to see that limx→0sinx−xx2=0,
but I can't figure out for the life of me how to argue without using L'Hospital or Taylor. Any ideas?
Answer
In THIS ANSWER, I used the integral definition of the arcsine function to show that for 0≤x≤π/2, we have the inequalities
xcos(x)≤sin(x)≤x
Using the trigonometric identity 1−cos(x)=2sin2(x/2), we see from (1) that
−2x(sin2(x/2)x2)⏟→14≤sin(x)−xx2≤0
Applying the squeeze theorem to (2) yields the coveted limit
limx→0sin(x)−xx2=0
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