Thank you! Evaluate the limit Solution Take the limit: lim_(x->infinity) (-4 x-8

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Thank you!

Explanation / Answer

Take the limit: lim_(x->infinity) (-4 x-8 x^2+10 x^3)/(10-10 x-8 x^3) Indeterminate form of type infinity/infinity. Using L'Hospital's rule we have, lim_(x->infinity) (-4 x-8 x^2+10 x^3)/(10-10 x-8 x^3) = lim_(x->infinity) (( d(-4 x-8 x^2+10 x^3))/( dx))/(( d(10-10 x-8 x^3))/( dx)): = lim_(x->infinity) (2+8 x-15 x^2)/(5+12 x^2) Indeterminate form of type infinity/infinity. Using L'Hospital's rule we have, lim_(x->infinity) (2+8 x-15 x^2)/(5+12 x^2) = lim_(x->infinity) (( d(2+8 x-15 x^2))/( dx))/(( d(5+12 x^2))/( dx)): = lim_(x->infinity) (-5/4+1/(3 x)) The limit of a constant is the constant: The limit of a sum is the sum of the limits: = -5/4+1/3 (lim_(x->infinity) 1/x) The limit of a quotient is the quotient of the limits: The limit of a constant is the constant: = -5/4+1/(3 (lim_(x->infinity) x)) The limit of x as x approaches infinity is infinity: Answer: | | = -5/4

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