Let\'s prove that lim_t right arrow 0 |sin(t)| = 0. Let D = [1, 0] and B be a po
ID: 1721368 • Letter: L
Question
Let's prove that lim_t right arrow 0 |sin(t)| = 0. Let D = [1, 0] and B be a point on the unit circle with centre A = [0, 0]. Let C be the projection of B onto the line AD, and let t be the angle angleBAD in radians, as shown below. The length of the straight line segment BC can be expressed in terms of t as (we require lengths to be positive or zero, so use an absolute value). The length of the circular arc BD can be expressed in term of t as By comparing these lengths, we get the inequality 0 less than or equal to |sin (t)| less than or equal to |t| for all t. Since then by the pinching theorem Note: the Maple syntax for |sin(x)| is abs (sin (x)).Explanation / Answer
According to the triangle ,
sin(t) = BC/AB
Since radius of the circle is 1 unit , therefore AB = 1
=> BC = sin(t)
BD = Radius * Angle(BAD)
BD = 1 * t
BD = t
Therefore , the value of limit is 0
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