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If $f(x+y)=f(x)+f(y)$ for all real values of $x,y$. Given $f(1)=1$

How to evaluate $$\lim_{x \to 0} \frac{2^{f(\tan x)}-2^{f(\sin x)}}{x^{2}f(\sin x)}$$

gt6989b
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    Could you share your own attempt at the solving the problem? – Zach466920 Jun 24 '15 at 16:22
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    Is $f$ known to be continuous as well? Not sure it can be done unless $f$ is continuous (and the $f(x)=1$ for all $x$. – Thomas Andrews Jun 24 '15 at 16:29
  • @ThomasAndrews $f(x)=x$ for all $x$ – Sepideh Abadpour Jun 24 '15 at 16:36
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    Whoops, that's what I meant. Brain burp. @sepideh – Thomas Andrews Jun 24 '15 at 16:53
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    $f(1)=1$ is a red herring, it's not really needed. The same proof (given in answer) will work if $f(1) = a$. However, f being continuous (or some weaker regularity such as bounded on some interval or being measurable) is crucial. For a discontinuous solution to Cauchy functional equation I suspect that the limit won't exist. – dioid Jun 24 '15 at 17:44
  • @dioid Could you please tell me why continuity of $f$ and $f(x+y)=f(x)+f(y)$ yields to $f(x)=x$? – Sepideh Abadpour Jun 24 '15 at 17:57
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    $f(n+1)=f(n)+f(1)$ for $n$ natural gives $f(n)=nf(1)$ by induction and similarly $f(1) = f(m/m) = f(1/m) + \ldots f(1/m) = mf(1/m)$ gives $f(1/m) = 1/m f(1)$ and then $f(r) = rf(1)$ for $r$ rational. Then f continuous gives $f(x) = xf(1)$. In your proof $f(1) = a$ for any real number $a$ will give the same limit. – dioid Jun 24 '15 at 18:19
  • @dioid Regarding $f(x+y)=f(x)+f(y)$ and by induction we can say for $k$ as an integer we have $f(kx)=kf(x)$ and because $f(1)=1$ we will have $f(k)=k$ but that's just for integer numbers. How can I prove that $f(x)=x f$ for $x$ as a real number? – Sepideh Abadpour Jun 24 '15 at 18:23
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    In general $f(rx) = rf(x)$ for $r$ a rational number if $f(x+y) = f(x) + f(y)$ by considering $m/m = 1/m + \ldots 1/m$ (with $m$ terms in the sum). Now if f is continuous then $f(r) = rf(1)$ for r rational then by considering limits $f(x) = xf(1)$ since f is continuous. – dioid Jun 24 '15 at 18:31
  • OK @dioid how about irrational ones? Can we say that continuity plus $f(r)=rf(1)$ for rational numbers is enough to say that we have $f(i)=if(1)$ for i as an irrational number? and why? Can we be sure that of all irrational numbers is rational numbers? and so then use continuity to say that $f(x)=xf(1)$ is true for irrational numbers? – Sepideh Abadpour Jun 24 '15 at 18:34
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    Yes, for any irrational number $x$ you have a sequence $r_n$ of rational numbers such that $r_n \to x$ and this gives (by continuity of $f$) that $f(x) = f(\lim r_n) = \lim f(r_n) = \lim r_n f(1) = xf(1)$. – dioid Jun 24 '15 at 18:36
  • @dioid thanks for your answer. It was really helpful. I don't know how to add to your privileges. – Sepideh Abadpour Jun 24 '15 at 18:39

2 Answers2

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Hint: If $f(x + y) = f(x) + f(y)$, then $$f(0 + 1) = f(0) + f(1).$$ Therefore, $f(0) = f(0 + 1) - f(1)$ and thus $f(0) = 1 - 1 = 0$. So $f(0) = 0$.

Decaf-Math
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We have $f(2x)=f(x)+f(x)=2f(x)$ and $f(3x)=f(2x)+f(x)=3f(x)$

Also we have $f(1)=f(0+1)=f(0)+f(1)\Rightarrow f(0)=0$

Also $f(0)=0=f(x+(-x))=f(x)+f(-x)\Rightarrow f(-x)=-f(x)$

Then $f(-2x)=f(-x)+f(-x)=-f(x)-f(x)=-2f(x)$
and $f(-3x)=f(-2x-x)=f(-2x)+f(-x)=-2f(x)-f(x)=-3f(x)$

So by induction we will have: $$\forall k\in\mathbb Z:f(kx)=kf(x)$$ so we will have $f(\frac{m}{m})=mf(\frac{1}{m})\Rightarrow f(\frac{1}{m})=\frac{1}{m}f(1)\;\forall m\in\mathbb Z$
Then for rational numbers we will have $$f(r)=f(\frac{m}{n})=mf(\frac{1}{n})=m\times\frac{1}{n}f(1)=\frac{m}{n}f(1)=rf(1)\; \forall r\in\mathbb Q$$
We now that for any irrational $x$ we have a sequence of rational numbers $r_n$ such that $\lim_{n\to \infty}r_n=x$ so from the continuity of $f$ we will have: $$f(x)=f(\lim_{n\to\infty}r_n)=\lim_{n\to\infty}f(r_n)=\lim_{n\to\infty}r_nf(1)=f(1)\lim_{n\to\infty}r_n=xf(1)$$ So because the proof is true for rational and irrational numbers we will have: $$f(x)=xf(1)=x\times 1=x\quad\forall x\in\mathbb R$$

So if $f$ is continuous then we have $f(x)=x$ for all $x$ so: $$\lim_{x \to 0} \frac{2^{f(\tan x)}-2^{f(\sin x)}}{x^{2}f(\sin x)}=\lim_{x\to 0}\frac{2^{\tan x}-2^{\sin x}}{x^2 \sin x}=\lim_{x\to 0}\frac{e^{\tan x\, \ln2}-e^{\sin x\, \ln2}}{x^2\sin x}$$

From the Maclaurin series we have $x\to 0:\sin x\simeq x$

Also from the Maclaurin series we have $x\rightarrow0\Rightarrow e^x\simeq1+x$
Substituting $x$ by $\sin x\,\ln2$ and $\tan x\,\ln2$ because both of them tend t0 $0$ when $x\rightarrow 0$ yields: $$=\lim_{x\to 0}\ln2\frac{\tan x-\sin x}{x^3}$$
Then if we write Maclaurin series up to the third derivative for $\sin x$ and $\tan x$ we have:
$x\to 0:\sin x\simeq x-\frac{x^3}{6}\quad,\quad \tan x\simeq x+\frac{x^3}{3}$
so we will have $$=\lim_{x\to 0}\ln2\frac{x+\frac{x^3}{3}-(x-\frac{x^3}{6})}{x^3}=\lim_{x\to 0}\frac{\ln2}{2}\frac{x^3}{x^3}=\frac{\ln2}{2}$$

Vim
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Sepideh Abadpour
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