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Given $f:\mathbb{R}\rightarrow \mathbb{R}$ differentiable continuously, and $g:\mathbb{R}\rightarrow \mathbb{R} \ s.t.\lim_{x \rightarrow 0}g(x)=0$ $$\text{Prove:} \lim_{x \rightarrow 0} \frac{f(x+g(x))-f(g(x))}{x}=f'(0) $$

$f'(x)$ is continuous $\Rightarrow\lim_{x \rightarrow 0}f'(g(x))=f'(\lim_{x \rightarrow 0}g(x))=f'(0) $

By definition of the derivative: $\lim_{x \rightarrow 0} \frac{f(x+0)-f(0)}{x}=f'(0)$

How can I continue to get the result?

Any help appreciated.

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  • $\begingroup$ Follows directly from Newton's formula doesn't it, i mean the difference quotient. $\endgroup$ – samjoe Mar 9 '17 at 13:39
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    $\begingroup$ I am sure there must be a way to do this using first principles. But if you can use L'hospital's rule, then the given limit (of the form $\frac{0}{0}$) is simplified to $$\lim_{x \rightarrow 0} \frac{f^{'}(x+g(x))(1 + g^{'}(x))-f^{'}(g(x))g^{'}(x)}{1}=f'(0)$$ $\endgroup$ – Shraddheya Shendre Mar 9 '17 at 13:41
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    $\begingroup$ @ShraddheyaShendre I see it now, but why is that ok not to place $0$'s in $g(x)$? Is it a number to me ? And also, it's not given that $g$ is differentiable. $\endgroup$ – Itay4 Mar 9 '17 at 13:48
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Since $f'$ is continuous, we may write first :

$$f(g(x)+x)-f(g(x))=\int_{g(x)}^{g(x)+x}f'(t)\,dt$$

then, for all $x\neq 0$ :

$$\frac{f(g(x)+x)-f(g(x))}{x}-f'(0)=\frac{1}{x}\int_{g(x)}^{g(x)+x}\left(f'(t)-f'(0)\right)\,dt$$

Now, given $\epsilon>0$, there exists $\eta>0$ such that :

$$|t|\le\eta\implies|f'(t)-f'(0)|\le\epsilon$$

Since $\lim_{x\to0}g(x)=\lim_{x\to0}\left(g(x)+x\right)=0$, there exists $\alpha>0$ such that :

$$|x|\le\alpha\implies\left(|g(x)|\le\eta\quad\mathrm{and}\quad|g(x)+x|\le\eta\right)$$

Hence, as soon $|x|\le\alpha$ :

$$\left|\frac{f(g(x)+x)-f(g(x))}{x}-f'(0)\right|\le\frac{1}{|x|}|x|\epsilon=\epsilon$$


Another proof

For all nonzero $x$, there exists $c_x\in\left(g(x),g(x)+x\right)$ or $\left(g(x)+x,g(x)\right)$ (depending on whether $x>0$ or $x<0$) such that :

$$f\left(g(x)+x\right)-f\left(g(x)\right)=\left((g(x)+x)-g(x)\right)\,f'(c_x)=x\,f'(c_x)$$

By squeeze theorem, we have $\lim_{x\to0}c_x=0$ because $\lim_{x\to0}g(x)=0$.

Hence, since $f'$ is continuous : $\lim_{x\to0}f'(c_x)=f'(0)$; in other words :

$$\lim_{x\to0}\frac{f\left(g(x)+x\right)-f\left(g(x)\right)}{x}=f'(0)$$

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  • $\begingroup$ Thank you for your answer, and sorry for not mentioning, but I'm looking for a solution without integrals. $\endgroup$ – Itay4 Mar 9 '17 at 14:01
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    $\begingroup$ @Itay4: Ok, here is another proof, but be aware that is depends on the axiom of choice (for each $x\neq0$, we choose some $c_x$ such that ...). $\endgroup$ – Adren Mar 9 '17 at 14:12
  • $\begingroup$ Why is that ok not to relate to the case when $x=0$? $\endgroup$ – Itay4 Mar 9 '17 at 14:21
  • $\begingroup$ First you don't want to divide by $0$ ... Second, it's ok because we have the mean value theorem which asserts that if $a<b$ and if $f:[a,b]\to\mathbb{R}$ continuous on $[a,b]$ and differentiable on $(a,b)$, then there exists $c\in(a,b)$ such that $f(b)-f(a)=(b-a)f'(c)$. $\endgroup$ – Adren Mar 9 '17 at 14:26

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