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This question is concerning a recent post: Find the follwing limit if $f$ is differentiable

Before looking at any of the answers posted I solved this problem using L'Hospital's rule, which results in the correct answer, as was noted by two other users. However, I'm unsure why the method is unsound. The individuals who used L'Hospital's rule in their solution had their answer downvoted. Specifically, @Paramanand Singh stated "L'Hospital's Rule does not apply here because we don't know if f(x) is differentiable at points other than x=0."

I went back to the textbook I was taught Calculus from, and it states L'Hospital's rule to be applicable if the functions f(x) and g(x) (making up the numerator and denominator of f(x)/g(x)) are differentiable on the interval (a,b), along with the necessary conditions mentioned in this post: When to Use L'Hôpital's Rule

My question, in response to @Paramanand Singh's statement: because we are given that f(x) is differentiable at x=0 (in the first link), would it be valid if I assumed f(x) is differentiable on the interval (0-ε,0+ε) and then applied L'Hospital's rule?

Mainly, could someone give me a more detailed explanation as to why L'Hospital's rule should not have been used?

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  • $\begingroup$ Good to see my comment being discussed. It would have been better if someone asked me directly by responding to my comment and I would be more than willing to clarify. There are lot of finer points in calculus which many beginners don't understand and conditions for applicability of L'Hospital's Rule are not often verified before its application. Sometimes experienced users also make mistakes in this area. $\endgroup$ – Paramanand Singh Jun 11 '17 at 13:27
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Just because a function is differentiable in a single point it doesn't have to differentiable on an interval containing that point.

For example, let $f(x) = 0$ if $x \in \mathbb Q$ and $f(x) = x^2$ if $x \in \mathbb R \setminus \mathbb Q$. Then $f'(0) = 0$, but $f'(x)$ isn't defined for any other $x$.

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  • $\begingroup$ I see what you are saying. That is why I was trying to be "clever" by defining an infinitesimally small interval using epsilon. I see why it doesn' hold up, there are too many other possible scenarios. So in this particular case, L'Hospital's rule should not be used because I have no information about the interval (which is also why my epsilon idea is also invalid)? $\endgroup$ – Preston Roy Jun 10 '17 at 21:21
  • $\begingroup$ You have understood it correctly. Also, there are no "infinitesimally small interval" in $\mathbb R$. $\endgroup$ – md2perpe Jun 10 '17 at 21:28

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