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I see a lot of calculus texts stating direct substitution is a form of evaluation for a limit. Maybe I'm missing something because, to me, direct substitution only shows the value of a function $f(x,y)$ for a given value of $(x,y)$. Can we necessarily assume that the limit of the function around $(x,y)$ also converges to that value?

Maybe I need to see a proof to understand if someone can point me in the right direction.

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the crux of it is that it depends on continuity. by definition, a continuous function's limit at $(a,b)$ is just the value of the function at that point. once you've deduced continuity (or manipulated the formula to the point where it's evident), you can just substitute the numbers in –  citedcorpse Jun 11 '13 at 10:47
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This is valid precisely if $f$ is continuous. –  Hagen von Eitzen Jun 11 '13 at 10:48

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Direct substitution works for limits only if the value of the function at the limiting point is defined, in other words that there is no discontinuity at that point. In fact the very definition of continuity for a two variable function is that the limit is just equal to the value when substituted:

A function $f(x,y)$ is continuous at $(a,b)$ if $$\lim_{(x,y)\to(a,b)}f(x,y)=f(a,b).$$

Graphically this has the same meaning as in single variable calculus, i.e. that the function doesn't have holes or jumps.

enter image description here

Otherwise, direct substitution fails and you should use other techniques such as employing polar coordinates.

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fyi: There is another user with the same name. –  Bill Dubuque Aug 16 at 14:29
    
Oh I'm so sorry I didn't know that so please can you edit my name so that it doesn't create this problem? (I can't edit it myself) –  Math Gems Aug 16 at 17:04

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