Thank you! Evaluate the following limits. If needed, enter \'infinity\' for infi
ID: 3342168 • Letter: T
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Thank you!
Evaluate the following limits. If needed, enter 'infinity' for infinity and '-infinity' for -infinity.Explanation / Answer
Take the limit: lim_(x->infinity) sqrt(-9-5 x^3+x^4)/(-6+3 x^2) Simplify radicals, sqrt(-9-5 x^3+x^4)/(-6+3 x^2) = sqrt((-9-5 x^3+x^4)/(36-36 x^2+9 x^4)): = lim_(x->infinity) sqrt((-9-5 x^3+x^4)/(36-36 x^2+9 x^4)) Using the power law, write lim_(x->infinity) sqrt((-9-5 x^3+x^4)/(36-36 x^2+9 x^4)) as sqrt(lim_(x->infinity) (-9-5 x^3+x^4)/(36-36 x^2+9 x^4)): = sqrt(lim_(x->infinity) (-9-5 x^3+x^4)/(36-36 x^2+9 x^4)) Indeterminate form of type infinity/infinity. Using L'Hospital's rule we have, lim_(x->infinity) (-9-5 x^3+x^4)/(36-36 x^2+9 x^4) = lim_(x->infinity) (( d(-9-5 x^3+x^4))/( dx))/(( d(36-36 x^2+9 x^4))/( dx)): = sqrt(lim_(x->infinity) (x (-15+4 x))/(36 (-2+x^2))) Factor out constants: = 1/6 sqrt(lim_(x->infinity) (x (-15+4 x))/(-2+x^2)) Indeterminate form of type infinity/infinity. Applying L'Hospital's rule we have, lim_(x->infinity) (x (-15+4 x))/(-2+x^2) = lim_(x->infinity) (( d(x (-15+4 x)))/( dx))/(( d(-2+x^2))/( dx)): = 1/6 sqrt(lim_(x->infinity) (-15+8 x)/(2 x)) Factor out constants: = sqrt(lim_(x->infinity) (-15+8 x)/x)/(6 sqrt(2)) Indeterminate form of type infinity/infinity. Using L'Hospital's rule we have, lim_(x->infinity) (-15+8 x)/x = lim_(x->infinity) (( d(-15+8 x))/( dx))/(( dx)/( dx)): The limit of a constant is the constant: Answer: | | = 1/3
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