If f is differentiable at 1, and the limit of f(1+h)/h as h->0 is 5, what is f(1)?

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If f is differentiable at 1, and the limit of f(1+h)/h as h->0 is 5, what is f(1)?

Mathematics
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` f is differentiable at 1` so \[\Large \lim_{h\to 0}\frac{f(1+h)-f(1)}{h}\] exists and it is defined based on the limit definition of the derivative
so \[f'(1) = 5-\lim_{h \rightarrow 0}\frac{f(1)}{h}\]

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Other answers:

\[\Large \lim_{h\to 0}\frac{f(1+h)}{h}=5\] \[\Large \lim_{h\to 0}\frac{f(1+h){\LARGE \color{red}{-0}}}{h}=5\] \[\Large \lim_{h\to 0}\frac{f(1+h){\LARGE \color{red}{-f(1)}}}{h}=5\] so we see that \(\Large f(1) = 0\)
so my step would be invalid?
it's valid, I just don't see where to go with it
oh ok. And you knew to put in that f(1) as 0 simply because the equation resembled the derivative limit?
I saw it match up with the limit definition. Well almost match up. It was just missing the `-f(1)` part
so f(x) = 0 and f'(x) = 5?
f ' (1) = 5 and f(1) = 0 we can't say anything about f(x) or f ' (x) in general
ok, thank you
IF it was \[\Large \lim_{h\to 0}\frac{f(x+h)-f(x)}{h}=5\] then we can say f ' (x) = 5
ok got it

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