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I know this must be wrong, but I am confused as to where the mathematical fallacy lies. Here is the 'proof':

$$f '(x) = \lim_{ h\to0}\frac{f(x+h)-f(x)}{h}$$

L'Hopital's Rule (Previous limit was 0/0):

$$ f '(x) = \lim_{ h\to 0}\frac{f '(x+h)-f '(x)} {1} $$

Plugging in h:

$$ f '(x) = f '(x+0)-f '(x) $$

Simplifying:

$$ f '(x) = 0 $$

I'm assuming my application of L'Hopital's rule is fallacious, but it evaluates to indeterminate so isn't L'Hopital's rule valid?

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2 Answers 2

up vote 38 down vote accepted

Taking the derivative with respect to $h$ gives:

$$f'(x) = \lim_{h \rightarrow 0} \frac{f'(x + h)}{1}$$

Since $f(x)$ is constant with respect to $h$.

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You differentiated the numerator with respect to $x$ but the denominator with respect to $h$.

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