How do we go from $\lim_{h\to0}=\frac{f(a+h)-f(a)}{h}$ to $\lim_{h\to0}=\frac{f(a+h)-f(a)-Lh}{h}=0$
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$$ \lim_{h \to 0} \frac{f(a+h) - f(a)}{h} = L \Leftrightarrow \lim_{h \to 0} \left(\frac{f(a+h) - f(a)}{h} - L\right) = 0 \Leftrightarrow \lim_{h \to 0} \frac{f(a+h) - f(a) - Lh}{h} = 0 $$
Clement Yung
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Just as an additional question, how do we know L = f'(a) in order for it to equal 0? – mathgeek101 Oct 31 '19 at 01:16
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If you can find such an $L$ such that the equation is satisfied, then it's necessary and sufficient that $L = f'(a)$. – Clement Yung Oct 31 '19 at 01:17
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Is it possible for there to be a proof for that? – mathgeek101 Oct 31 '19 at 01:19
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There's nothing to prove. The definition of $f'(a)$ is precisely $\lim_{h \to 0} \frac{f(a+h) - f(a)}{h}$. – Clement Yung Oct 31 '19 at 01:21
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AH I see it now, thanks! – mathgeek101 Oct 31 '19 at 01:23