Is there a rule that says you have to simplify equations before evaluating them? Would $y=\frac{x}{x}$ at $x=0$ be $1$ or undefined, since without reducing it, you'd divide by $0$. I know the equation can simplify to $y=1$, but I thought simplification just made the equation easier to work and was not necessary. It still should yield the same results. I don't no if this is a bad example as the equation reduces to a constant, but the question would still hold up for any power of $x$ divide by $x$. For example, $y=\frac{x^2}{x}$ still has the same issue. At $x=0$ does it evaluate to $0$ or undefined. Sorry if this is a dumb question. Just one of my musings that I've wondered about.
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$\begingroup$ $x\rightarrow 0$ not $x=0$ $\endgroup$– E.H.EOct 8, 2015 at 20:10
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1$\begingroup$ $y=\frac{x}{x}$ would be undefined at $x=0$...The graph would have a hole at $x=0$ and everywhere else $y=1$. $\endgroup$– randomgirlOct 8, 2015 at 20:10
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$\begingroup$ Hint: Simplifyng $y=\frac{x^2}{x}$ you have $y=x$ if $x\ne 0$ .... but, as a function this is different from $y=\frac{x^2}{x}$ . $\endgroup$– Emilio NovatiOct 8, 2015 at 20:10
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1$\begingroup$ As long as $x \neq 0$ right @EmilioNovati ? $\endgroup$– randomgirlOct 8, 2015 at 20:11
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$\begingroup$ @randomgirl if y=x/x has a whole in it @ x=0, then is y=x/x not the same as y=1? $\endgroup$– HoopdadyOct 8, 2015 at 20:12
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2 Answers
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The proper term to use is "removable discontinuity". You're correct that the expression equals 1 in pretty much every case - still, 0 is not a part of that function's domain. The function you gave is equivalent to this: $$f(x)=1, x\neq0$$
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I believe more precisely than the "division by zero" that you mention in the title, your question is about $0/0$. To study it, it is better to turn to the plane: rather than $z=f(x)=x/x$, let's consider $z=g(x,y)=x/y$. Then you can define different ways to reach $0/0$. If you follow the $x=0$ line with decreasing $y$, you'll get $\lim_{y\rightarrow 0} g(0,y) = 0$. If you follow the line $x=y$, then $\lim_{y\rightarrow 0} g(y,y) = 1$. And if you follow the curve $y=x^2$, you obtain $\lim_{x\rightarrow 0^\pm} g(x,x^2) =\lim_{x\rightarrow 0^\pm} 1/x = \pm\infty$ !
Thus the function $f(x,y)=x/y$ does not have a continuous extension in $(0,0)$. The ratio $0/0$ is really 'undefined' for this reason. But if you know that you'll be approaching $0/0$ along a definite path (e.g. $y=x$), it is valid to simplify as you will obtain a function which is the continuous extension of the one you started with.
