using the definition of derivative, prove that the limit as x approaches 0 of (s
ID: 3082071 • Letter: U
Question
using the definition of derivative, prove that the limit as x approaches 0 of (sin(x))/x=1 and (cos(x)-1)/x=0.Explanation / Answer
The limit is indeed 1. This is easy to see from a fairly well-known identity: lim sin(x) / x = 1 x->0 The function sin(x) / x is continuous where x is not 0, so the limit of the reciprocal, i.e. x / sin(x), is the reciprocal of 1 (i.e. 1 itself). The above identity is proved using a geometric argument and squeeze theorem, and is required to use L'Hospital's rule with trigonometric functions (so using L'Hospital's rule to prove the limit is cheating ;) EDIT: If you want that proof of the limit of sin(x) / x, check this out: This is supposed to be a segment of a radius 1 circle, with an angle of x radians, and in it, we construct an isosceles triangle, and outside it, we construct a right-angled triangle. The segment's area will be simply half the angle times the radius squared, so just (1/2)x * 1^2 = (1/2)x. The isosceles triangle's area, according the (1/2)absinC formula for area of a triangle, will be (1/2)1*1*sin(x) = (1/2)sin(x). The outsides triangle will have a leg length of 1 (the radius) and tan(x) (the other, tangential leg). So the area will be (1/2) * 1 * tan(x) = (1/2)tan(x) The area of the segment will always be between the other two areas, i.e: (1/2)sin(x) < (1/2)x < (1/2)tan(x) sin(x)Related Questions
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