find a nice smooth function $\lim_{x\to\infty} (f(x^2)-f(x))=1$, $\lim_{x\to\infty} (f(x)f(\frac{1}{x}))=1$ and $\lim_{x\to-\infty} (f(x)-f(x^2))=1$? can you find a nice real and smooth function where $\lim_{x\to\infty} (f(x^2)-f(x))=1$, $\lim_{x\to\infty} (f(x)f(\frac{1}{x}))=1$ and $\lim_{x\to-\infty} (f(x)-f(x^2))=1$?
I've figured a function that follows the first limit.$$\lim_{x\to\infty} (f(x^2)-f(x))=1$$
I first tried logarithms and they didn't work. but it seemed close.
$$\ln(x^2)-\ln(x)=\ln(x)$$
So next I tried logarithms of logarithms. and got $\ln(2)$ so I divided my function by $\ln(2)$.
$$\frac{\ln(\ln(x^2))}{\ln(2)}-\frac{\ln(\ln(x))}{\ln(2)}=1$$
because for all x this is true $\frac{\ln(\ln(x))}{\ln(2)}$ is a solution if we ignore the other two limits, the vertical asymptote at 1 and not being defined for all real numbers.



$$\lim_{x\to\infty} (f(x)f(\frac{1}{x}))=1$$
The second limit I also found a solution for which is $f(x)=x^n$ where n is any real number because $x^n\times\frac{1}{x}^n=1$



$$\lim_{x\to-\infty} (f(x)-f(x^2))=1$$
for the last one I made sure to use the first on and reordered the terms to make it work and got $-\frac{\ln(\ln(-x))}{\ln(2)}$.
$$-\frac{\ln(\ln(-x))}{\ln(2)}+\frac{\ln(\ln(-x^2))}{\ln(2)}=1$$
even though I could find an example of a function that followed on of the limits, none of them followed all three limits and two of them weren't even defined for all real numbers. So can you find a function that follows all three of these limits, whose domain and range are the real numbers, and are smooth meaning to me continuous and differentiable everywhere.
 A: No, you cannot.
$\textbf{Claim;}$ There is no smooth function that satisfies both
$$1) \text{ }\text{ }\text{ }\lim_{x \rightarrow \infty} f(x^2)-f(x) =1$$
$$2) \text{ }\text{ }\text{ }\lim_{x \rightarrow \infty} f(x)f(\frac{1}{x}) = 1$$
We will show that such a function cannot be differentiable at $0$.
$\textbf{Proof;}$ Suppose there is such a function, then we can find an $M > 0$ so that for $x > M$ we have $f \neq 0$ (by (2)),
$$3)\text{ }\text{ }\text{ }\frac{1}{2}+f(x) \leq f(x^2) \leq \frac{3}{2}+f(x)$$
and
$$4)\text{ }\text{ }\text{ } \frac{\frac{1}{2}}{f(x)} \leq f(\frac{1}{x}) \leq \frac{\frac{3}{2}}{f(x)}$$
Choose $v > \max\{1,M\}+1 > 1$ and set $f(v) = s$
By $3)$ we have
$$5)\text{ }\text{ }\text{ }s +\frac{1}{2}n \leq  f(v^{2^n}) \leq s +\frac{3}{2}n $$
thus by $4)$ and $5)$
$$6)\text{ }\text{ }\text{ }\frac{\frac{1}{2}}{s +\frac{3}{2}n} \leq f(\frac{1}{v^{2^n}}) \leq \frac{\frac{3}{2}}{s +\frac{1}{2}n}.$$
By continuity of $f$ and $6)$ we must have $f(0) = 0$. Now for all $n \in \mathbb{N}$
$$7)\text{ }\text{ }\text{ } f'(0) = \lim_{n \rightarrow \infty}\frac{f(\frac{1}{v^{2^n}})-f(0)}{\frac{1}{v^{2^n}}} \geq \liminf_{n \rightarrow \infty}\frac{\frac{1}{2}}{s +\frac{3}{2}n}v^{2^n} = \infty.$$
This is a contradiction.
