'Tough' Limits Problem using the 'sign' function

[Flopper]

Hi! I'm taking a Calculus II course and am struggling with the following problem.. I think there is a piece of information about limits I'm missing here.

The sign function sgn (x) is a defined as follows:

x/(abs)x.. if x is not equal to 0
0.. if x is equal to 0

Use the sign function to define two functions f and g whose limits as x-->0 do not exist, but such that

a.) lim [f(x) + g(x)] does exist
x-->0
b.) lim f(x)*g(x) does exist
x-->0

Could this simply be f(x)=x for negative numbers and g(x)=x^2 for positive numbers or the reverse (doesn't matter).. intuitively I feel this couldn't be the right answer.

Need help..

 

[soroban]

Re: 'Tough' Limits Problem..

Hello, Flopper!

The sign function $\displaystyle \text{sgn}(x)$ is defined as follows:

. . $\displaystyle \text{sgn}(x) \:=\:\begin{Bmatrix}\frac{x}{|x|} & \;\;\; & \text{if }x\,\neq\,0 \\ \\ \\ 0 & \;\;\; &\text{if }x\,=\,0\end{Bmatrix}$

Use the sign function to define two functions $\displaystyle f$ and $\displaystyle g$
whose limits as $\displaystyle x\,\to\,0$ do not exist, but such that:

$\displaystyle a)\;\lim_{x\to0}[f(x)\,+\,g(x)]\,$ does exist

$\displaystyle b)\;\lim_{x\to0}f(x)\cdot g(x)\,$ does exist

Your thinking is in the right direction . . .


One idea comes to mind: .$\displaystyle \begin{array}{ccc}f(x) & = & |\text{sgn}(x)| \\ \\ \\ g(x) & = & -|\text{sgn}(x)| \end{array}$


We see that: \(\displaystyle \begin{array}{c}f(x) \text{ returns }+1\text{ (or 0)} \\ \\ \\
g(x) \text{ returns }-1\text{ (or 0)}\end{array}\; \text{ and }\lim_{x\to0}\text{ does not exist.}\)


(a) For $\displaystyle x\,\neq\,0:\;f(x)\,+\,g(x)\;=\;\text{sgn}(x)\,-\,\text{sgn}(x) \;=\;0$

Therefore: $\displaystyle \:\lim_{x\to0}\bigg[f(x)\,+\,g(x)\bigg] \;=\;\lim_{x\to0}[0] \;=\;0$


(b) For $\displaystyle x\,\neq\,0:\;f(x)\cdot g(x)\;=\;\bigg[\text{sgn}(x)\bigg]\cdot\bigg[-\text{sgn}(x)\bigg] \;=\;(1)(-1)\;=\;-1$

Therefore: $\displaystyle \:\lim_{x\to0}\bigg[f(x)\cdot g(x)\bigg] \;=\;\lim_{x\to0}[-1] \;=\;-1$