Originally Posted by

**HallsofIvy** I recommend you go back and re-read the problem. What you wrote here, makes no sense. $\displaystyle \lim_{x\to 0} x+ b= b$ so you make that limit equal to "L" by taking b= L. But then b= L can be any number. Did the problem say something about "continuous"?

$\displaystyle \lim_{x\to 0} F(x)= \infty$, $\displaystyle \lim_{x\to 0} xF(x)= 0$. So F must go to infinity as x goes to 0 but not as fast as x itself. How about 1 over x to any power less than 1? Say, $\displaystyle 1/\sqrt{x}= x^{-1/2}$?

$\displaystyle \lim_{x\to \infty} F(x)= 0$, $\displaystyle \lim_{x\to 0^-}F(x)= 3$, $\displaystyle \lim_{x\to 0^+}F(x)= 1$

As you say, do this as two separate problems. First find a function that goes to 0 as x goes to infinity and goes to 1 as x goes to 0. There are many such functions. A fraction with denominator of higher degree than numerator but with constant terms in numerator and denominator the same with work. Then find a function that goes to 3 as x goes to 0 (for x< 0 but if you find one that works for x> 0, just replace x with -x).