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proving $\frac{1}{x}$ by definition

$$\left(\frac{1}{x}\right)'=\lim_{h \to 0} {\frac{1}{x+h}-\frac{1}{x}\over h}=\lim _{h \to 0} {\frac{x-x-h}{(x+h)x}\over h}=\lim _{h \to 0} {\frac{-h}{(x+h)x}\over h}=\lim _{h \to 0} {\frac{-1}{(x+h)x}}$$

now can I say that as $h \to 0$ so $(x+h)=x$ shouldn't ${\frac{-1}{(x+h)x}}$ be continuous for that? or because it is not approaching $-\frac{1}{0}$ that is ok?

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    $\begingroup$ Yes you do need continuity but rational functions are continuous everywhere except at the points where the denominator is zero. Both of your arguments are the same in this case. $\endgroup$ Jun 23 '15 at 16:30
  • $\begingroup$ @CameronWilliams Which is to say that rational functions are continuous, they are just not always $\Bbb R\to \Bbb R$. $\endgroup$
    – Arthur
    Jun 23 '15 at 17:07
  • $\begingroup$ It's like to say that $1\over x$ is continuous at $\mathbb{R^+}$? $\endgroup$
    – gbox
    Jun 23 '15 at 17:26
  • $\begingroup$ @Arthur good point since they're not defined at their discontinuities anyway. $\endgroup$ Jun 23 '15 at 18:48
  • $\begingroup$ You have proved that the function has a derivative (at some domain), your question is not clear. Are you asking about continuity of the function? If true, see:math.stackexchange.com/questions/354632/… $\endgroup$
    – NoChance
    Jan 15 '16 at 4:52

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