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Goal: To show that
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If we try to evaluate the limit of sine of theta divided by theta as theta approaches 0 with algebraic techniques, we will be unsuccessful. Instead, we turn to a geometric analysis and the Squeeze Theorem. The method that is to be employed is
- to set up three expressions where for two of them the limit can be found directly
- structure the expresssions into inequalities, where the third expression (the one where we cannot find the limit directly) is forced to be between the other two
- once the limits at the low end and high end are found and shown to be equal we will be able to state with certainty that the third expression must have the exact same limit.
In this case, we expect the limit to be 1, as shown in the following graph and numeric table:
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In the following sketches, the base of each region has the same length. The area of triangle ABC is found by
. [For a full explanation: click.] The area of the sector A’B’C’ is found by
. The area of triangle A”B”C” is found by
. From the sketches, we can see that the area of triangle ABC is clearly greater than the area of sector A’B’C’, which is clearly greater than the area of triangle A”B”C”, no matter the measure of the angle (in radians). If angle theta is reduced to a 0 radian measure, all area would be lost, thus the relationship of the areas is that the red triangle is greater than or equal to the blue sector which has area greater than or equal to the green triangle.
Thus, we can set up the following inequality:
[Note: we use "greater than or equal to" rather than "strictly greater than" because if the angle is 0 radians, then the area of each is zero.]
Let the length of the base be defined as 1 unit, yielding:
Multiply through the inequality by
, yields:
Since
, we substitute, yielding
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We are attempting to prove
Take the limit of each piece as theta approaches 0:
Since,
and
, then we can say
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Thus, the limit of sine of theta divided by theta as theta approaches 0 must equal 1. This is a classic application of the Squeeze Theorem. The "trick" in this limit was to set up the correct areas and to structure the three algebraic representations into inequalities. The key was to find two expressions with the same limit and would force the third to lie between the other two. Graphically, we see that in the "neighborhood" of theta equalling zero (theta not equal to zero) the graph of sine of theta (in blue) is entirely between the graph of cosine of theta (green) and the graph of y = 1 (red).
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Once the limits at the low end and high end were found and shown to be equal we can state with certainty that the third expression must have the exact same limit. Hence,
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are
important because they play roles in establishing the derivative
of the sine and cosine functions (and thereby helping define
the derivatives of all the other trigonometric functions).