compute limx->inf (x^3+5x)/(2x^3-x^2+4) Solution lim_(x->infinity) (x^3+5 x)/(2

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lim_(x->infinity) (x^3+5 x)/(2 x^3-x^2+4) Indeterminate form of type infinity/infinity. Using L'Hospital's rule we have, lim_(x->infinity) (x^3+5 x)/(2 x^3-x^2+4) = lim_(x->infinity) (( d(5 x+x^3))/( dx))/(( d(4-x^2+2 x^3))/( dx)): = lim_(x->infinity) (3 x^2+5)/(6 x^2-2 x) Indeterminate form of type infinity/infinity. Using L'Hospital's rule we have, lim_(x->infinity) (3 x^2+5)/(6 x^2-2 x) = lim_(x->infinity) (( d(5+3 x^2))/( dx))/(( d(-2 x+6 x^2))/( dx)): = lim_(x->infinity) (3 x)/(6 x-1) Factor out constants: = 3 (lim_(x->infinity) x/(6 x-1)) Indeterminate form of type infinity/infinity. Using L'Hospital's rule we have, lim_(x->infinity) x/(6 x-1) = lim_(x->infinity) (( dx)/( dx))/(( d(-1+6 x))/( dx)): = 1/2

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