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Gives a function $f(x, y, z)$ and a positive number $\epsilon .$ In each exercise, show that there exists a $\delta>0$ such that for all $(x, y, z)$.$$\sqrt{x^{2}+y^{2}+z^{2}}<\delta \Rightarrow|f(x, y, z)-f(0,0,0)|<\epsilon$$$$f(x, y, z)=x y z, \quad \epsilon=0.008$$

$x=r \sin (\theta) \cos (\phi)$$y=r \sin (\theta) \sin (\phi)$$z=r \cos (\theta)$

Calculus 3

Chapter 14

Partial Derivatives

Section 2

Limits and Continuity in Higher Dimensions

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12:15

In calculus, partial derivatives are derivatives of a function with respect to one or more of its arguments, where the other arguments are treated as constants. Partial derivatives contrast with total derivatives, which are derivatives of the total function with respect to all of its arguments.

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Gives a function $f(x, y, …

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s in this case, You complete Delta on Need Square. Really? That's one. The reasoning is that it's follows s we can see that after the origin is your square implicit square pursuits were zero. So the absolute difference between F and itself at the origin just absolutely do that. What is that? Since it is a great people to zero use of its summons X squared y squared likewise being called a zero. So we only need to look a no es is a square of the rude Excluding y squared is that square I wish I mean x squared plus y squared plus c squared in the course by assumption Uh, this is strictly what's indelible squared which by assumption initially here, Delta dude up Sign that the squared is that absolutely And that is what the result we want to achieve with this question

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