I'm having problems understanding one of the rules of the algebra of functional limits, specifically this one:
$\lim_{x \to a}[f(x)\cdot g(x)]=\lim_{x \to a} f(x)\cdot \lim_{x \to a}g(x)$
I have a counter example in mind, which I know doesn't hold for some reason, just now sure why. If we let
$f(x)=x-a$ $,$ $g(x)=a$
then
$\lim_{x \to a}f(x)=0$ , $\lim_{x \to a}g(x)=a$ therefore $\lim_{x \to a} f(x)\cdot \lim_{x \to a}g(x)=0$
but
$\lim_{x \to a}[f(x)\cdot g(x)]=\lim_{x \to a}[a\cdot(a-x)]=\lim_{x \to a}[a^2-ax]=a$
Would love any insight into where I've gone wrong, thank you.
The expression :- $\lim_{x \to a}[f(x)\cdot g(x)]=\lim_{x \to a} f(x)\cdot \lim_{x \to a}g(x)$ holds only if $\lim_{x \to a}f(x)$ and $\lim_{x \to a} g(x)$ . Exist separately.
For a basic counter example. take the sequence $f(x)=x$ and $g(x)=\sin(\frac{1}{x})$.
Then the limit as $x\to 0$ of the product is $\lim_{x\to 0}x\sin(\frac{1}{x})=0$
But the limit $\lim_{x\to 0}g(x) $ does not exist.
Otherwise consider an even basic example:-
$f(x)=\{\begin{align} 1\,,\text{x is rational}\\ 0, \text{x is irrational}\\ \end{align}$
And take
$g(x)=\{\begin{align} 0\,,\text{x is rational}\\ 1, \text{x is irrational}\\ \end{align}$
Then the limit of f and g do not exist for any point in $\mathbb{R}$ but the product function $f(x)g(x)$ is just the $0$ function ,i.e it takes the value $0$ $\forall\,x\in\mathbb{R}$ .
So the limit of the product function should just be $0$.
If you want counter-examples from sequences then take $a_{n}=(-1)^{n}$ and $b_{n}=(-1)^{n+1}$