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needhelp07Best ResponseYou've already chosen the best response.0
\[\lim_{x \rightarrow \infty}(\frac{ x ^{3}1 }{ x ^{4}+1 })\]
 one year ago

shiningBest ResponseYou've already chosen the best response.1
easily consider the highest rate of varaiabe seperately in nominator and denominator lim x^3/x^4 = 0
 one year ago

needhelp07Best ResponseYou've already chosen the best response.0
how did u do that??
 one year ago

shiningBest ResponseYou've already chosen the best response.1
i said you should just put the highest power of varaiable instead of up and down (because the limit approaches infinity) so\[\lim \frac{ x^{3} }{ x^{4}? }\]=0 because x^4 > x^3
 one year ago

needhelp07Best ResponseYou've already chosen the best response.0
what happens to 1 and +1 ?? is it cancelled out?? sorry if i have many questions
 one year ago

shiningBest ResponseYou've already chosen the best response.1
you shoul eliminate them because in the limits of fraction like this type constants have no touchable influence
 one year ago

ShanksBest ResponseYou've already chosen the best response.0
you can understand this by taking x^2 and x^4 common as follows: => \[\lim_{x \rightarrow \infty} \left( \frac{ x^{2} ( 1 \frac{ 1 }{x^{2}}) }{ x^{4} ( 1+ \frac{ 1 }{x^{4}}) } \right)\] Further strike out and reduce it to \[\lim_{x \rightarrow \infty} \left( \frac{( 1 \frac{ 1 }{x^{2}}) }{ x^{2} ( 1+ \frac{ 1 }{x^{4}}) } \right)\] Obviously 1/x^2 and 1/x^4 when x> infinity will be = 0 So inside bracket fraction becomes 1/1. and 1/x^2 = 0 when x> infinity Thereby, given lim = 0
 one year ago
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