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Show that the function in Example 6 has limit 0 along every straight line approaching (0,0)

0

Calculus 3

Chapter 14

Partial Derivatives

Section 2

Limits and Continuity in Higher Dimensions

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Lectures

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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Show that the function in …

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Show that the function $f(…

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Show that the functions ha…

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Consider the function $f(x…

06:23

Show that $f(x, y)$ has no…

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Show that the indicated li…

in this question, we will start by first considering vertical line at X equals zero. So our clam it When x and y approaches Iran zero long X equals zero and for our function to excess squish waas over export four plus white boards to if we substitute actually quit zeroing dysfunction We will have toe but the place you propose toe Why over the report Fourth Plus why push to so it will be equal to zero Now we can consider the non vertical line through the Iran zero. So the equation off any line parting through the U. N zero is off the form. Why equals m X So let's substituting our limit when x and y up coaches zero and zero along Why equal Emmett and next Sorry. Oh, the function off Any line putting through the reindeer These in four Why equal Imex? Not in key eso this limit for our function. If off x and y will be to over two x square Why over x or four plus why over two now we can substitute by equal m x so we will have to thanks Course three m over export four loss and those two exports toe. We can take them export to as a common factor. So toe exports three. I am over exposed to supply exports toe plus m course toe. Now can we can remove export to is exports toe So our solution will be equal toe X over exports toe plus and forced to Now we can substitute X in our function equals zero So we will have to see the solution will be zero So our limit for dysfunction along any line putting through zero n zero will be equal zero. Thank you.

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