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I have the function $f(x) = 1 + \frac{12x+4}{\left( x+1 \right)^2} \cdot \left( \frac{12}{12x+4} - \frac{2}{x+1} \right)$ (actually the derivate of another function, but that shouldn't matter). Since $12x+4=0$ for $x=\frac{-1}{3}$ I would have said the function value is undefined at that point - but Wolframalpha says it's 28! Now what I think might have happened is that WA secretly showed me the limit of $x\to\frac{-1}{3}$ and just didn't bother to tell me. Or, am I missing something else?

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What exactly did you type in WA? Obviously it canceled the $12x+4$! –  Blah May 5 '12 at 18:13
    
1 + (12x+4)/((x+1)^2) * (12/(12x+4) - 2/(x+1)), x=-1/3 –  Cubic May 5 '12 at 18:16
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That's OK. Removable singularities are for removing. –  André Nicolas May 5 '12 at 18:16
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-1 for Wolfram Alpha –  David Mitra May 5 '12 at 18:21
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I find it annoying that these details (if you consider it a detail) are usually glossed over without comment by teachers. –  copper.hat May 5 '12 at 18:28

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up vote 0 down vote accepted

if we consider function without multiplication ,yes it is undefined at $x=-1/3$ ,but we can remove it by multiplication ,so after we execute this operation we get $f(x)=1+12/(x+1)^2-(24*x+8)/(x+1)^3$,at $x=-1/3$ yes $f(-1/3)$=28

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Can I just do it like that? Granted, the original function is continuous at $-1/3$ so I can probably do that in this case, but are such things ok in general? –  Cubic May 5 '12 at 18:39
    
it depends,as others mentioned ,it is like removable point,like in limits,so you may do it in some cases,but actually i think that it requires some permision before do this –  dato datuashvili May 5 '12 at 18:41
    
i hope it would help you,good lucks –  dato datuashvili May 5 '12 at 18:52

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