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anonymous
 4 years ago
Horizontal asymptote of the function f(x)= 6x^3+17x^2528/2x^314x+321
anonymous
 4 years ago
Horizontal asymptote of the function f(x)= 6x^3+17x^2528/2x^314x+321

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anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[f(x) = 6x^3+17x^2528/2x^314x+321\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0can someone help me please

Callisto
 4 years ago
Best ResponseYou've already chosen the best response.0Horizontal asymptotes \[=f(x) = \lim_{x \rightarrow \infty} \frac{6x^3+17x^2528}{2x^314x+321}=...\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0yes thats the problem

Callisto
 4 years ago
Best ResponseYou've already chosen the best response.0Divide both the numerator and denominator by x^3

Callisto
 4 years ago
Best ResponseYou've already chosen the best response.0And \[\lim_{x \rightarrow \infty} \frac{1}{x} =0\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0what is the next step , i believe 3 is right

Callisto
 4 years ago
Best ResponseYou've already chosen the best response.0\[\lim_{x \rightarrow \infty} \frac{6x^3+17x^2528}{2x^314x+321}\]\[=\lim_{x \rightarrow \infty} \frac{\frac{6x^3+17x^2528}{x^3}}{\frac{2x^314x+321}{x^3}}\]\[=\lim_{x \rightarrow \infty} \frac{6+\frac{17}{x}\frac{528}{x^3}}{2\frac{14}{x^2}+\frac{321}{x^3}}\]Then take the limit
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