7
$\begingroup$

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?

$\endgroup$
  • 2
    $\begingroup$ Show that...? You seem to have omitted something. $\endgroup$ – Cameron Williams May 25 '15 at 16:50
  • $\begingroup$ oopse, fixed... $\endgroup$ – Nescio May 25 '15 at 16:51
  • 2
    $\begingroup$ Do you also want a limit on the left hand side of that expression? $\endgroup$ – Cameron Williams 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
  • 2
    $\begingroup$ 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
4
$\begingroup$

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

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ What does $C^1$ denote? $\endgroup$ – user26486 May 26 '15 at 0:55
0
$\begingroup$

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.

| cite | improve this answer | |
$\endgroup$
0
$\begingroup$

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.

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.