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lim ((1/c+h)-(1/c))/h h → 0
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\[\lim_{h \rightarrow 1} \frac{ ((\frac{ 1 }{ c+h })-(\frac{ 1 }{ c })) }{ h }\] To make the equation more clear.
sorry, as h-->0
put over common denominator c-c+h ______ = h/(c+h)h=1/(c+h) / h= 1/(c+h)h= 1/(c+0)0 = 1/0=+infinity (c+h)h
\[\lim_{h \to 1} \frac{ \dfrac{ 1 }{ c+h }-\dfrac{ 1 }{ c } }{ h }\\ \lim_{h \to 1} \frac{ \dfrac{ c }{ c(c+h) }-\dfrac{ c+h }{ c(c+h) } }{ h }\\ \lim_{h \to 1} \frac{ \dfrac{ c -c-h}{ c(c+h) }}{ h }\\ -\lim_{h \to 1} \frac{ \dfrac{ h}{ c(c+h) }}{ h }\\ -\lim_{h \to 1} \dfrac{1}{ c(c+h) }=\cdots \]
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