Squeeze-defining a function If for all real numbers $3>a>5/2>b>2$, there exist $e>0$ such that $x^b<f(x)<x^a$ for all $x>e$, must $f(x)=x^{5/2}$?
Followups:
If for all real numbers $3>a>5/2>b>2$, $x^b<f(x)<x^a$ for all $x$, must $f(x)=x^{5/2}$?
If for all real numbers $e>0$ and $3>a>5/2>b>2$, $x^{b-e}<f(x)<x^{a+e}$ for all $x$, must $f(x)=x^{5/2}$?
 A: Original question. No. The function $x^{5/2} (\ln x)^{k}$ (for any real $k$) is a counterexample. 
Follow-up question 1. The answer is yes. Suppose $x^b < f(x) < x^a$ for all $b < 5/2 < a$. Note that this hypothesis cannot hold for any $x \leq 1$, so we'll assume $x > 1$. 
Fix any $x > 1$. Then we have:
$$
b \ln x < \ln f(x) < a \ln x \ \ \implies \ \ b < \frac{\ln f(x)}{\ln x} < a, 
$$
for all $b < 5/2 < a$ (remember that $\ln x > 0$). Since $\frac{\ln f(x)}{\ln x} < a$ for all $a > 5/2$, it follows that $\frac{\ln f(x)}{\ln x} \leq 5/2$ (why?). We can similarly show $\frac{\ln f(x)}{\ln x} \geq 5/2$. Hence
$$
\frac{\ln f(x)}{\ln x} = \frac 52 \ \ \iff f(x) = x^{5/2} ,
$$
which is what we wanted. 
It is an interesting exercise to understand why the above argument fails in the case of the counterexample $x^{5/2} (\ln x)^k$. 
Followup question 2. If $x^{b-e} < f(x) < x^{a+e}$ for all $b < 5/2 < a$ and all $e > 0$, then we can show that the seemingly stronger condition $x^{b} < f(x) < x^{a}$ also holds for all $b < 5/2 < a$. (Exercise!) So this question is the same as the previous follow-up question. 
