<s>[INST] Difference quotient of differentiable functions $Q. \text { For a differentiable function } f : \mathbb { R } \rightarrow \mathbb { R } \text { define the differential quotient}$
\begin{array} { l }{ \left( D _ { x } f \right) ( h ) = \frac { f ( x + h ) - f ( x ) } { h } ; h > 0 } \\ { \text { consider numbers of the form } \hat { h } = h ( 1 + \epsilon ) \text { for a } } \\ { \text { fixed } \epsilon > 0 \text { and let } } \end{array}
\begin{array} { l } { e _ { 1 } ( h ) = f ^ { \prime } ( x ) - \left( D _ { x } f \right) ( h ) } \\ { e _ { 2 } ( h ) = \left( D _ { x } f \right) ( h ) - \left( D _ { x } f \right) ( \hat { h } ) } \\ { e ( h ) = e _ { 1 } ( h ) + e _ { 2 } ( h ) } \\ { \text { If } f ( x + \hat { h } ) = f ( x + h ) , \text { then, which are true, } } \end{array}
\begin{array} { l } { \text { (1) } e _ { 1 } ( h ) \rightarrow 0 \text { as } h \rightarrow 0 } \\ { \text { (2) } e _ { 2 } ( h ) \rightarrow 0 \text { as } h \rightarrow 0 } \\ { \text { (3) } e _ { 2 } ( h ) \rightarrow \epsilon f ^ { \prime } ( x ) / ( 1 + \epsilon ) \text { as } h \rightarrow 0 } \\ { \text { (4) } e(h)\rightarrow 0 \text { as } h \rightarrow 0 } \end{array}
My work: 
\begin{array} { l } { \text { Here } f \left( x + \hat h \right) = f ( x + h ) , \quad \hat h = h ( 1 + \epsilon ) , 
\quad \epsilon >0 \text { fixed. } } \\ { \qquad \begin{aligned} \Rightarrow D _ { x } f ( \hat h ) & = \frac { f ( x + \hat h ) - f ( x ) } { \hat h } \\ & = \frac { f ( x + h ) - f ( x ) } { h ( 1 + \epsilon ) } \\ & \rightarrow \frac { f ^ { \prime } ( x ) } { 1 + \epsilon } \text { as } h \rightarrow 0 \end{aligned} } \end{array}
\begin{array} { l } { \text { Also } h \rightarrow 0 \Rightarrow \hat h \rightarrow 0 } \\ { \begin{aligned} \Rightarrow & D _ { x } f ( \hat { h } ) = \frac { f ( x + \hat { h } ) - f ( x ) } { \hat h } \rightarrow f ^ { \prime } ( x ) \\ & \text { as } h \rightarrow 0 \\ & \text { Hence } \frac { f ^ { \prime } ( x ) } { 1 + \varepsilon } = f ^ { \prime } ( x ) \end{aligned} } \end{array}
$\Rightarrow \epsilon = 0 , \text { Where the mistake happened? }$
 [/INST] Not sure if this is what your looking for, but your last implication is incorrect since 
$\frac{f'(x)}{1+ \varepsilon} = f'(x)$ and $\varepsilon >0$ implies $$f'(x) = 0.$$
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