Question about an hypothesis in a limit with Hopital/Taylor

[Ozma]

Suppose [imath]f:\mathbb{R} \to \mathbb{R}[/imath] is such that [imath]f(0)=1[/imath] and [imath]f'[/imath] exists continuous in [imath]x=0[/imath]. Show that [imath]\lim_{x \to 0} (f(x))^{1/x}=e^{f'(0)}[/imath].

I proved this as follows: from Taylor's expansion of [imath]f[/imath] in [imath]x=0[/imath], it is [imath]f(x)=f(0)+f'(0)x+o(x)[/imath], but by hypothesis [imath]f(0)=1[/imath] and so it is [imath]f(x)=1+f'(0)x+o(x)[/imath]. Since [imath]1/x[/imath] is, in general, real because by hypothesis [imath]x \in \mathbb{R}[/imath], it follows that [imath]f(x)[/imath] must be positive due to the fact that power with real exponents are defined only for positive bases. Hence it is defined [imath]\log[(f(x))^{1/x}][/imath], so
[math]\lim_{x \to 0} (f(x))^{1/x}=\lim_{x \to 0} e^{\frac{1}{x}\log(f(x))}=\lim_{x \to 0} e^{\frac{1}{x}\log[1+f'(0)x+o(x)]}[/math]Since as [imath]x \to 0[/imath] it is [imath]f'(0)x+o(x) \to 0[/imath], from logarithm's Taylor expansion it is
[math]\lim_{x \to 0} e^{\frac{1}{x}\log[1+f'(0)x+o(x)]}=\lim_{x \to 0} e^{\frac{1}{x}[f'(0)x+o(x)]}=\lim_{x \to 0} e^{f'(0)+\frac{o(x)}{x}}=e^{f'(0)}[/math]However, it seems to me that I did not use the continuity of [imath]f'[/imath] in [imath]x=0[/imath]; my textbook solves this using Hospital's rule, and indeed it needs the continuity of [imath]f'[/imath] in [imath]x=0[/imath] because it ends up with [imath]\lim_{x \to 0} e^{\frac{f'(x)}{f(x)}}[/imath]. Is my solution correct as well and so the continuity of [imath]f'[/imath] in [imath]x=0[/imath] is not needed or I have made a mistake somewhere? If I recall correctly, Taylor's expansion only needs differentiability to the [imath]k[/imath]-th order, so in this case only to the first order, so continuity of [imath]f'[/imath] in [imath]x=0[/imath] should not be necessary. Am I right?

 

[Zermelo]

Suppose [imath]f:\mathbb{R} \to \mathbb{R}[/imath] is such that [imath]f(0)=1[/imath] and [imath]f'[/imath] exists continuous in [imath]x=0[/imath]. Show that [imath]\lim_{x \to 0} (f(x))^{1/x}=e^{f'(0)}[/imath].

I proved this as follows: from Taylor's expansion of [imath]f[/imath] in [imath]x=0[/imath], it is [imath]f(x)=f(0)+f'(0)x+o(x)[/imath], but by hypothesis [imath]f(0)=1[/imath] and so it is [imath]f(x)=1+f'(0)x+o(x)[/imath]. Since [imath]1/x[/imath] is, in general, real because by hypothesis [imath]x \in \mathbb{R}[/imath], it follows that [imath]f(x)[/imath] must be positive due to the fact that power with real exponents are defined only for positive bases. Hence it is defined [imath]\log[(f(x))^{1/x}][/imath], so
[math]\lim_{x \to 0} (f(x))^{1/x}=\lim_{x \to 0} e^{\frac{1}{x}\log(f(x))}=\lim_{x \to 0} e^{\frac{1}{x}\log[1+f'(0)x+o(x)]}[/math]Since as [imath]x \to 0[/imath] it is [imath]f'(0)x+o(x) \to 0[/imath], from logarithm's Taylor expansion it is
[math]\lim_{x \to 0} e^{\frac{1}{x}\log[1+f'(0)x+o(x)]}=\lim_{x \to 0} e^{\frac{1}{x}[f'(0)x+o(x)]}=\lim_{x \to 0} e^{f'(0)+\frac{o(x)}{x}}=e^{f'(0)}[/math]However, it seems to me that I did not use the continuity of [imath]f'[/imath] in [imath]x=0[/imath]; my textbook solves this using Hospital's rule, and indeed it needs the continuity of [imath]f'[/imath] in [imath]x=0[/imath] because it ends up with [imath]\lim_{x \to 0} e^{\frac{f'(x)}{f(x)}}[/imath]. Is my solution correct as well and so the continuity of [imath]f'[/imath] in [imath]x=0[/imath] is not needed or I have made a mistake somewhere? If I recall correctly, Taylor's expansion only needs differentiability to the [imath]k[/imath]-th order, so in this case only to the first order, so continuity of [imath]f'[/imath] in [imath]x=0[/imath] should not be necessary. Am I right?

If a function is differentiable, then it’s continious. ?

 

[Ozma]

@Zermelo In my solution I have not used the continuity of [imath]f'[/imath] at [imath]x=0[/imath] given by the hypothesis of the problem, nothing is said about the continuity of [imath]f[/imath]. Differentiability only assures the continuity of [imath]f[/imath], so I don't get what you are suggesting.

 

[Deleted member 4993]

f in x=0x=0x=0, it is f(x)=f(0)+f′(0)x+o(x)f(x)=f(0)+f'(0)x+o(x)f(x)=f(0)+f′(0)x+o(x),

By above (OP) you are claiming f'(0) exists ← you are assuming f(x) is continuous at x=0

 

[Ozma]

@Subhotosh Khan I agree with that, but the question was about whether the hypothesis about the continuity of the derivative of f in x=0 is necessary or not, not about the continuity of f in x=0 which I agree is necessary (because, as you say, it is implied by the differentiability of f).

 

[Zermelo]

@Subhotosh Khan I agree with that, but the question was about whether the hypothesis about the continuity of the derivative of f in x=0 is necessary or not, not about the continuity of f in x=0 which I agree is necessary (because, as you say, it is implied by the differentiability of f

Ooooh, now I understand your question, you’re asking about the continuity of f’ .
In this case, f’ doesn’t have to be continious if f is differentiable.

This statement of Taylor’s theorem states that the derivatives must be continious, so you can’t use Taylor’s expansion without f’ being continious.
But sometimes mathematicians simplify these statements, maybe there is a proof for Taylor’s theorem that needs n derivatives to exist (thus n-1 of them are continious), and the nth one doesn’t have to be continious. Not sure right now! Search the internet, if you find a statement of Taylor’s theorem which doesn’t need the last derivative to be continious, then you’ve prooved the same thing with less assumptions, congrats ?
But in general I think you shouldn’t worry about this stuff too much, usually we say that a function is n times continiously differentiable, even though the last derivative doesn’t have to be continious, it’s annoying to have to remember that little detail for every theorem

 

[Ozma]

@Zermelo Thank you for the help! Exactly, I am trying to confirm if the continuity of the last derivative used in the expansion with the little-o remainder is required. I will search and try to find out, or wait here if someone is so gentle to confirm.