Discover The Proof Behind The Limit Of Sin(Theta) / Theta As Theta Approaches 0 = 1 Using Taylor Series Expansion.

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

1

The limit of sin(theta) / theta as theta approaches 0 is equal to 1.

To see why this is the case, we can use the fact that sin(theta) can be written as a Taylor series expansion centered around 0:

sin(theta) = theta – (theta^3 / 3!) + (theta^5 / 5!) – (theta^7 / 7!) + …

Dividing both sides by theta and taking the limit as theta approaches 0, we get:

lim theta->0 sin(theta) / theta = [1 – (theta^2 / 3!) + (theta^4 / 5!) – (theta^6 / 7!) + …] / 1

As theta approaches 0, each term in the expansion approaches 0, except for the first term which approaches 1. Therefore, the limit of sin(theta) / theta as theta approaches 0 is equal to 1.

More Answers:
The Derivative Of A Constant Function: Why It’S Always 0
Mastering Calculus – The Power Rule For Easy Polynomial Function Derivatives
Evaluating (1 – Cos(Theta)) / Theta Using Trigonometric Limits And L’Hopital’S Rule

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