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across
 2 years ago
Best ResponseYou've already chosen the best response.0Anything can happen, but generally, you're approaching the limit point from negative infinity.

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

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

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

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

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

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

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

pradipgr817
 2 years ago
Best ResponseYou've already chosen the best response.0I have tried it differently dw:1347746473941:dw at left as h tends 0 x tends\[2^{}\]

pradipgr817
 2 years ago
Best ResponseYou've already chosen the best response.0so if we replace x by 2h and let h tends to 0 then there will be a different solution.

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

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