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haterofmath
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What hapeens if the limit is being taken from the left?
 2 years ago
 2 years ago
haterofmath Group Title
What hapeens if the limit is being taken from the left?
 2 years ago
 2 years 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.
 2 years 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?
 2 years ago

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

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

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

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

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

helder_edwin Group TitleBest ResponseYou've already chosen the best response.0
yes. if \(x\to2+\) then \(x>2\) so \(x2>0\)
 2 years 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\[2^{}\]
 2 years 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.
 2 years ago

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

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