Here's the question you clicked on:
haterofmath
What hapeens if the limit is being taken from the left?
Anything can happen, but generally, you're approaching the limit point from negative infinity.
so in a problem such as \[\lim_{x \rightarrow 2-}\frac{ x^2+4 }{ x-2 }\] howw would it work?
since \(x\to2-\) then \(x<2\) so \(x-2<0\)
this means that \[ \Large \lim_{x\to2-}\frac{x^2+4}{x-2}=-\infty \]
ok, so i won't have to plug anything in?
and if it were moving to the right it would be \[+\infty\]
well, pluging in you would have \[ \Large \frac{2^2+4}{2-2}=\frac{8}{0} \]
yes. if \(x\to2+\) then \(x>2\) so \(x-2>0\)
I have tried it differently |dw:1347746473941:dw| at left as h tends 0 x tends\[2^{-}\]
so if we replace x by 2-h and let h tends to 0 then there will be a different solution.
and what solutions that?
you can put it and find it.