limit as x approaches 0: 1-cosx/x
0
To evaluate the given limit, we can either use algebraic manipulation or L’Hopital’s rule. I will show both methods:
Algebraic Manipulation:
We can rewrite 1-cosx/x as (1-cosx) * (1/x). Then, we can use the trigonometric identity: 1-cosx = 2sin^2(x/2) to rewrite the expression as 2sin^2(x/2)/x. Now, we can further simplify by using the identity sin(x)/x = 1 for small values of x. Thus:
limit as x approaches 0: 1-cosx/x
= limit as x approaches 0: 2sin^2(x/2)/x (Using trigonometric identity)
= limit as x approaches 0: 2 * (x/2)^2 / (x/2) * (x/2) (Using half-angle identity for sine)
= limit as x approaches 0: 2 * (1/2)^2 (Canceling x term)
= 1
Therefore, the given limit is 1.
L’Hopital’s Rule:
To use L’Hopital’s Rule, we need to take the derivative of the numerator and denominator separately with respect to x:
limit as x approaches 0: 1-cosx/x
= limit as x approaches 0: (sinx)/(1) (Applying L’Hopital’s Rule)
= 0/1
= 0
Therefore, the limit using L’Hopital’s Rule is 0. However, this method does not work in this case because we would be dividing by 0/0 when taking the derivative of the numerator. Therefore, we must use algebraic manipulation to evaluate the limit.
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