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acrossBest ResponseYou've already chosen the best response.0
Anything can happen, but generally, you're approaching the limit point from negative infinity.
 one year ago

haterofmathBest ResponseYou've already chosen the best response.0
so in a problem such as \[\lim_{x \rightarrow 2}\frac{ x^2+4 }{ x2 }\] howw would it work?
 one year ago

helder_edwinBest ResponseYou've already chosen the best response.0
since \(x\to2\) then \(x<2\) so \(x2<0\)
 one year ago

helder_edwinBest ResponseYou've already chosen the best response.0
this means that \[ \Large \lim_{x\to2}\frac{x^2+4}{x2}=\infty \]
 one year ago

haterofmathBest ResponseYou've already chosen the best response.0
ok, so i won't have to plug anything in?
 one year ago

haterofmathBest ResponseYou've already chosen the best response.0
and if it were moving to the right it would be \[+\infty\]
 one year ago

helder_edwinBest ResponseYou've already chosen the best response.0
well, pluging in you would have \[ \Large \frac{2^2+4}{22}=\frac{8}{0} \]
 one year ago

helder_edwinBest ResponseYou've already chosen the best response.0
yes. if \(x\to2+\) then \(x>2\) so \(x2>0\)
 one year ago

pradipgr817Best ResponseYou've already chosen the best response.0
I have tried it differently dw:1347746473941:dw at left as h tends 0 x tends\[2^{}\]
 one year ago

pradipgr817Best ResponseYou've already chosen the best response.0
so if we replace x by 2h and let h tends to 0 then there will be a different solution.
 one year ago

haterofmathBest ResponseYou've already chosen the best response.0
and what solutions that?
 one year ago

pradipgr817Best ResponseYou've already chosen the best response.0
you can put it and find it.
 one year ago
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