Prove the limit of 1/sqrt(3x+7) = 1/5 as x approaches 6 (using epsilon-delta) (ILIEKMATHPHYSICS)

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This video references the book "Introduction to Real Analysis" by Bartle and Sherbert (Fourth Edition). For more details related to the types of exercises covered in this video, see section 4.1.

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Prove the limit of 1/sqrt(3x+7) = 1/5 as x approaches 6 (using epsilon-delta) (ILIEKMATHPHYSICS)

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Комментарии
Автор

haha nice. Just getting into proofs and i love your channel

Artemis
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It can be done more easily. The function f(x)=1/sqrt(3x+7) is obviously continuous in a neighbourhood of 6. But then lim_{x->6}f(x) =
1/sqrt(3(lim_{x->6})+7) = 1/sqrt(3*6+7)=1/5. The argument relies on the fact that lim_{x->y}f(x) = f(lim_{x->y}x) when f i contiuous at y.

kennethvalbjoern
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How convenient that you got a |x-6|. If you had had, say |x-9| you would do |x - 6 -3| and apply triangle inequality to x - 6 and 3?

radadadadee