# derivative $1 \over x$ -proof

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?

• 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. Jun 23 '15 at 16:30
• @CameronWilliams Which is to say that rational functions are continuous, they are just not always $\Bbb R\to \Bbb R$. Jun 23 '15 at 17:07
• It's like to say that $1\over x$ is continuous at $\mathbb{R^+}$?
– gbox
Jun 23 '15 at 17:26
• @Arthur good point since they're not defined at their discontinuities anyway. Jun 23 '15 at 18:48
• 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/… Jan 15 '16 at 4:52