Good example showing why limits must exist in limit product rule

Question

I'm looking for a way to show my calc 1 students not to use the limit laws without knowing that the individual limits exists. I could use $$\lim_{x\to 0} x^{2} \sin(1/x),$$

but by doing it wrong, one still gets the right answer, which is 0. So I'm looking for an example where splitting up the limit would actually give the wrong answer. I think $\lim_{x\to 0} x \csc(1/x)$ ought to work, but I'd like an even simpler example, if possible.

Answer 1

One can say that for $x$ to +infinity on $y=\sqrt{x^2-4x}$ that $\sqrt{x^2-4x}=\sqrt{x^2(1-4/x^2)}=x\sqrt{1-4/x^2}=x$ because the square root term goes to 1. Therefore the slant asymptote is $y=x$ This is an example of the incorrect use of limit laws since the first term, $x$ has a limit that does not exist for $x$ to infinity. There is a slant asymptote for $x$ to infinity that has a slope of $1$ but also a nonzero $b$ value. I want YOU to figure that out since you are a Calc 1 teacher :)

Answer 2

Maybe this is the kind of thing you are looking for:

Define the sequence $x_n=(-1)^n$.

"Therefore", we have that:

$1=\lim x_n^2"="(\lim x_n)^2.$

"So", $\lim x_n=1$ or $-1$.

Also, you could use $x$ and $\frac{1}{x}$ if they think that the argument "$0$.anything=$0$" is valid.