Proving The Limit Of Sin(Theta)/Theta As Theta Approaches 0 To Be Equal To 1: Using L’Hopital’S Rule And The Squeeze Theorem

lim theta->0 sin(theta) / theta = ___________________

1

The limit of sin(theta)/theta as theta approaches zero is equal to 1.

To prove this, we can use L’Hopital’s rule or the squeeze theorem. Here’s how:

Using L’Hopital’s rule:
We can rewrite the expression as (sin(theta))/theta, which is an indeterminate form of type 0/0, meaning both the numerator and denominator approach zero as theta approaches zero. Applying L’Hopital’s rule, we take the derivative of both the numerator and the denominator with respect to theta. The derivative of sin(theta) is cos(theta) and the derivative of theta is 1. So, we get:
lim theta->0 (sin(theta))/theta = lim theta->0 (cos(theta))/1 = cos(0)/1 = 1

Using the squeeze theorem:
We know that 0 <= sin(theta) <= theta for all theta in the first quadrant. This is because sin(theta) is the height of the red curve in a unit circle (centered at the origin), which is always less than or equal to the length of the arc connecting the two points on the circle where the arc is centered at the origin, and the angle of the arc is theta. Therefore, 0 <= sin(theta)/theta <= 1 for theta > 0.

Taking the limit as theta approaches zero of the inequalities above, we get:
lim theta->0 0 <= lim theta->0 sin(theta)/theta <= lim theta->0 1
0 <= lim theta->0 sin(theta)/theta <= 1 By the squeeze theorem, since the limits on the left and right are both equal to 0, we conclude that: lim theta->0 sin(theta)/theta = 1

Therefore, the limit of sin(theta)/theta as theta approaches zero is equal to 1.

More Answers: