Saturday, January 3, 2015
Unizor - Trigonometry - Lim (SIN(x)/x)
This lecture combines three different subjects, geometry, trigonometry and a theory of limits, into one extremely important theorem.
Consider an angle x, measured in radians, tends to zero, while remaining positive (staying positive is not really necessary for a theorem but makes it easier to explain the proof). Then the ratio of its sine to its value tends to 1, that is
lim[x→0] sin(x)/x = 1
Consider a unit circle around the origin of coordinates O intersecting a positive direction of the X-axis at point P.
Since we are talking in this theorem about an angle decreasing to zero while remaining positive, we need to consider only angles in the first quadrant.
Choose a point A on a unit circle, connect it with a center O and let angle ∠POA measure x radians. Connect point A with point P. Consider a triangle ΔAOP.
Let's compare the area of this triangle and the area of a sector of our unit circle between radius OP and radius OA. Obviously, the area of a sector, that includes the area of a triangle and an additional piece between a chord AP and an arc of a circle, is greater than the area of a triangle. Let's compare them quantitatively.
To determine an area of a triangle ΔAOP, drop a perpendicular from point A to OP, its base on OP being point Q. Obviously, AQ is an altitude in a triangle ΔAOP. The length of this altitude is an ordinate of point A and, therefore, by definition of an sine of an angle, equals to sin(x). The base of our triangle is a radius OP which is equal to 1. Therefore,
area (ΔAOP) = sin(x)/2
Area of a sector AOP that measures x radians is smaller than the area of an entire unit circle, which can be considered as a sector of 2π radians, by a factor of x/2π. The area of a unit circle equals to π·12=π. Therefore,
area (sector AOP) = (x/2π)·π = x/2
So, sin(x) is less than x.
Let's consider another pair of geometric figures. Extend radius OA beyond point A and draw a perpendicular to radius OP at point P. Point of intersection of this perpendicular with an extension to OA is point R. Consider a triangle ΔROP. Obviously, its area is greater than the area of a sector AOP we considered before because it completely includes it in itself. Let's quantify this inequality.
Triangle ΔROP is a right triangle with one cathetus OP equal to 1 and a ratio of another cathetus PR to this OP being a tangent of an angle measured x radians. Therefore, the area of triangle ΔROP equals to tan(x)/2 and x less than tan(x).
Putting together both inequalities, we come up with
sin(x) less than x less than tan(x).
Since we consider angle ∠POA to be a small positive angle measured x radians, all components of these two inequalities are positive. Let's divide all of them by a positive value of sin(x). The result is
1 less than x/sin(x) less than tan(x)/sin(x).
Since, by definition, tan(x)=sin(x)/cos(x), this inequality can be written as
1 less than x/sin(x) less than 1/cos(x)
1 greater than sin(x)/x greater than cos(x)
The last remaining step of the proof is to use a theorem from the theory of limits that, if a sequence is bounded from below by one sequence and from above by another, and both bounding sequences tend to the same limit, then the sequence "squeezed in-between them" tends to the same limit (see Algebra - Limits - Problems 3 of this course). Now, as angle ∠POA decreases to zero (that is, x→0), the upper bound of our ratio sin(x)/x remains constant 1, the lower bound of this ratio is cos(x), which at point x=0 also equals to 1 and, therefore, tends to 1 as x decreases to zero. Therefore, our ratio, "squeezed between" a constant 1 from above and variable cos(x) that also tends to 1, has a limit of 1:
lim[x→0] sin(x)/x) = 1
End of Proof.
Graphical interpretation of this limit is that the graph of function y=sin(x), as positive x gets closer and closer to zero, becomes more and more like the graph of function y=x, a straight line bisecting the first and third quadrant, while remaining just below this straight line.
Incidentally, as x approaches zero, ratio tan(x)/x also tends to 1 because it's not much different from sin(x)/x since tan(x)=sin(x)/cos(x) and cos(x) is almost equal to 1 for small x. Graphically, this limit means that the graph of function y=tan(x) behaves like the graph of function y=x in the positive vicinity of point x=0, while remaining just above it.