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mathcalculus
 one year ago
Best ResponseYou've already chosen the best response.1i got 3/5 but it's wrong.

mathcalculus
 one year ago
Best ResponseYou've already chosen the best response.1@kropot72 do you know this?

mathcalculus
 one year ago
Best ResponseYou've already chosen the best response.1@satellite73 any ideas?

hihihii
 one year ago
Best ResponseYou've already chosen the best response.0if the denominator grows bigger than the numerator, then the limit is 0

mathcalculus
 one year ago
Best ResponseYou've already chosen the best response.1oh so we don;t need to divide by x? or do anything to it?

mathcalculus
 one year ago
Best ResponseYou've already chosen the best response.1thanks! :) @hihihii

mathstudent55
 one year ago
Best ResponseYou've already chosen the best response.1\[\lim_{x \rightarrow \infty} \dfrac{3x + 2}{3x^2  5x + 8}\] \[ =\lim_{x \rightarrow \infty} \dfrac{ x^2 \left( \frac{3}{x} + \frac{2}{x^2} \right) } {x^2 \left( 3  \frac{5}{x} + \frac{8}{x^2} \right) }\] \[ = \lim_{x \rightarrow \infty} \dfrac{ \frac{3}{x} + \frac{2}{x^2} } { 3  \frac{5}{x} + \frac{8}{x^2} } \] \[ = \dfrac{ 0 + 0} { 3  0 + 0 } \] \(= 0\)

mathcalculus
 one year ago
Best ResponseYou've already chosen the best response.1thanks u @mathstudent55
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