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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'Hôpital's Rule (The previous limit was $\frac{0}{0}$):

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

Plugging in $h$:

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

Simplifying:

$$ f '(x) = 0 $$

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

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2 Answers 2

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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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  • $\begingroup$ Ah, and similarly, taking the derivative with respect to $x$ gives $\frac{f'(x+h) - f'(x)}{0}$, which would be invalid. At the time that I posted the question I wasn't familiar enough to Leibniz's notation to realize there's an implicit variable hidden in the apostrophe, but thanks for the clarification! $\endgroup$ Commented Oct 12, 2018 at 17:53
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You differentiated the numerator with respect to $x$ but the denominator with respect to $h$.

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