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I'm trying to answer the following question:

Let $f$ be continuously differentiable in all of $\mathbb{R}$ and let $g:\mathbb{R}\to\mathbb{R}$ be a function satisfying $\lim_{x\to0}g\left(x\right)=0$. Show that $$\lim_{x\to0} \frac{f\left(x+g\left(x\right)\right)-f\left(g\left(x\right)\right)}{x}=f'(0)$$

At first I though it would be pretty simple triangular inequality from the definition of a limit, but it doesn't really work out for me...

let $\epsilon>0$, we can find $\delta$ such that $$\left|x\right|<\delta\implies\left|\frac{f\left(x\right)-f\left(0\right)}{x}-f'\left(0\right)\right|<\frac{\epsilon}{2}$$ But as $g\left(x\right)\to0$ we can find $\delta_{g}$ such that

$$\left|x\right|<\delta_{g}\implies\left|g\left(x\right)\right|<\frac{\delta}{2}$$

Hence for $\left|x\right|<\min\left\{ \delta_{g},\frac{\delta}{2}\right\}$ we have that $x+g\left(x\right)<\delta$, and

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

Any direction?

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    $\begingroup$ Show that...? You seem to have omitted something. $\endgroup$ May 25 '15 at 16:50
  • $\begingroup$ oopse, fixed... $\endgroup$
    – Nescio
    May 25 '15 at 16:51
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    $\begingroup$ Do you also want a limit on the left hand side of that expression? $\endgroup$ May 25 '15 at 16:52
  • $\begingroup$ Oh wow... looks like I'm already too tried.. Thank you very much $\endgroup$
    – Nescio
    May 25 '15 at 16:54
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    $\begingroup$ Well you can use Lagrange's mean value theorem to say that whenever $g(x)\neq 0$ $$f(x+g(x))-f(g(x))=xf'(\xi_x),\ \text{ with }\xi_x\in[x-|g(x)|,x+|g(x)|]$$ and then use continuity of $f'$. And then, adjust the case when $g(x)$ is frequently $0$ in a nighbourhood of $0$. $\endgroup$
    – user228113
    May 25 '15 at 17:10
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Since $f(x)$ is $C^1$ you can use the mean value theorem. Namely $$\frac{f\left(x+g\left(x\right)\right)-f\left(g\left(x\right)\right)}{x} = f'(h(x))$$ Here $h(x)$ is between $g(x)$ and $x + g(x)$. Note that as $x$ goes to zero so does $h(x)$. So since $f'$ is continuous, as $x$ goes to zero $f'(h(x))$ goes to $f'(0)$, and thus you have $$\lim_{x \rightarrow 0} \frac{f\left(x+g\left(x\right)\right)-f\left(g\left(x\right)\right)}{x} = f'(0)$$

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  • $\begingroup$ What does $C^1$ denote? $\endgroup$
    – user26486
    May 26 '15 at 0:55
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Taylor's theorem is often a useful tool in this regard.

Taylor's theorem gives $f(z) = f(y) + f'(y)(z-y) + \int_y^z (f'(t)-f'(y))dt$, and so, for $x \neq 0$ \begin{eqnarray} {f(x+g(x)) -f(g(x)) \over x} -f'(0) &=& { f'(g(x)) x + \int_{g(x)}^{x+g(x)} (f'(t)-f'(g(x)))dt\over x}-f'(0) \\ &=& f'(g(x))-f'(0)+ { \int_{g(x)}^{x+g(x)} (f'(t)-f'(g(x)))dt\over x} \end{eqnarray}

Let $\epsilon>0$ and choose $r>$ such that $|f'(x)-f'(0)| < \epsilon $ for $x\in B(0,r)$. Now choose $\delta \le r$ such that $|g(x)| < r $ for $|x| < \delta$.

Then the above gives \begin{eqnarray} \left|{f(x+g(x)) -f(g(x)) \over x} -f'(0) \right| &\le& |f'(g(x))-f'(0)|+ | { \int_{g(x)}^{x+g(x)} (f'(t)-f'(g(x)))dt\over x} | \\ &\le& \epsilon + { \int_{g(x)}^{x+g(x)} |f'(t)-f'(g(x))|dt\over x} \\ &\le& \epsilon + \epsilon { \int_{g(x)}^{x+g(x)} dt\over x} \\ &=& 2 \epsilon \end{eqnarray}

Hence we have the desired result.

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If we use the chain rule and Taylor expansions at $x=0$ $$f(x+g(x))=f(g(0)\big)+x \left(g'(0) f'(g(0))+f'(g(0))\right)+O\left(x^2\right)$$ $$f(g(x))=f(g(0))+x g'(0) f'(g(0))+O\left(x^2\right)$$ $$f(x+g(x))-f(g(x))=x f'(g(0))+O\left(x^2\right)$$ Now, since $\lim_{x\to0}g\left(x\right)=0$, then the result.

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