como faço para resolver a alternativa E???

e)  f '(x) = lim _h→0 [ 1 / (x+ h) – 1 / x]  / h

      _{^{f '(x) =} lim} h→0  [ (x – x –h) / (x+ h). x ] / h

     f '(x) = lim _h→0 [ – h / (x +h) . x. h ] =  lim _h→0 [–h / h] . [1 /  (x+ h). x] =

= (– 1 ) . [ 1 /  x . x ]  =  – 1 /  ( x )^2

Sendo \(f(x) = {1 \over x}\), o limite é:

\(\Longrightarrow \underset{h \to 0} \lim \,\, {f(x+h) - f(x) \over h} = \underset{h \to 0} \lim \,\, {1 \over h} \Big (f(x+h) - f(x) \Big )\)

\(\Longrightarrow \underset{h \to 0} \lim \,\, {f(x+h) - f(x) \over h} = \underset{h \to 0} \lim \,\, {1 \over h} \Big ( {1 \over x+h} - {1 \over x} \Big )\)

\(\Longrightarrow \underset{h \to 0} \lim \,\, {f(x+h) - f(x) \over h} = \underset{h \to 0} \lim \,\, {1 \over h} \Big ( {x-(x+h) \over x(x+h)} \Big )\)

\(\Longrightarrow \underset{h \to 0} \lim \,\, {f(x+h) - f(x) \over h} = \underset{h \to 0} \lim \,\, {1 \over h} \Big ( {-h \over x(x+h)} \Big )\)

\(\Longrightarrow \underset{h \to 0} \lim \,\, {f(x+h) - f(x) \over h} = \underset{h \to 0} \lim \,\,-{1 \over x(x+h)}\)

Substituindo \(h=0\), o resultado final é:

\(\Longrightarrow \underset{h \to 0} \lim \,\, {f(x+h) - f(x) \over h} =-{1 \over x(x+0)}\)

\(\Longrightarrow \fbox {$ \underset{h \to 0} \lim \,\, {f(x+h) - f(x) \over h} =-{1 \over x^2} $}\)