AJ L. answered • 06/24/23
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Patient and knowledgeable Calculus Tutor committed to student mastery
Looks hard, but luckily, you can just use direct substitution:
limt->0 cos(π/√[19-3sec(2t)])
= cos(π/√[19-3sec(2·0)])
= cos(π/√[19-3sec(0)])
= cos(π/√(19-3))
= cos(π/√(16))
= cos(π/4)
= √2/2
Since the function evaluated at t=0 is f(0)=√2/2 which is defined, the limit of the function as t approaches 0 exists, and f(0) equals the value of the limit at t=0, then the function is continuous at t=0.
Roger R.
tutor
The conclusion "Therefore, the function is continuous at t = 0 since the limit exists." is false. When you plug in t = 0 to calculate the limit, you already use/ASSUME that the composite function is continuous at t = 0.
In your logic, the function f(x) = 0 if x is rational, f(x) = 1 if x is irrational, is continuous at x = 0 since you can plug in and calculate f(0) = 0.
You only proved that the composite function is defined at t = 0. It is a major theorem in calculus that the composition of continuous functions is continuous.
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06/25/23
AJ L.
tutor
You’re right, I rushed that ending.
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06/25/23
AJ L.
06/24/23