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haterofmath
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What hapeens if the limit is being taken from the left?
 one year ago
 one year ago
haterofmath Group Title
What hapeens if the limit is being taken from the left?
 one year ago
 one year ago

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across Group TitleBest 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

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

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

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

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

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

pradipgr817 Group TitleBest 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

pradipgr817 Group TitleBest 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

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

pradipgr817 Group TitleBest ResponseYou've already chosen the best response.0
you can put it and find it.
 one year ago
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