lim [(x^6 + 3x^5 + 4)^(1/6) -x]
x->infinity
The answer is 0.5. Please tell me how to do this. Thanks.
x->infinity
The answer is 0.5. Please tell me how to do this. Thanks.
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To keep this from getting too painful with conjugate-related tricks...
lim(x→∞) [(x^6 + 3x^5 + 4)^(1/6) - x]
= lim(x→∞) [(x^6 * (1 + 3/x + 4/x^6))^(1/6) - x]
= lim(x→∞) [x (1 + 3/x + 4/x^6)^(1/6) - x]
= lim(x→∞) x * [(1 + 3/x + 4/x^6)^(1/6) - 1]
= lim(x→∞) [(1 + 3/x + 4/x^6)^(1/6) - 1] / (1/x)
= lim(t→0+) [(1 + 3t + 4t^6)^(1/6) - 1] / t, letting t = 1/x.
Applying L'Hopital's Rule (since this is of the form 0/0) yields
lim(t→0+) [(1 + 3t + 4t^6)^(-5/6) * (3 + 24t^5) - 0] / 1
= lim(t→0+) (3 + 24t^5)/(1 + 3t + 4t^6)^(5/6)
= 3/1
= 3.
I hope this helps!
lim(x→∞) [(x^6 + 3x^5 + 4)^(1/6) - x]
= lim(x→∞) [(x^6 * (1 + 3/x + 4/x^6))^(1/6) - x]
= lim(x→∞) [x (1 + 3/x + 4/x^6)^(1/6) - x]
= lim(x→∞) x * [(1 + 3/x + 4/x^6)^(1/6) - 1]
= lim(x→∞) [(1 + 3/x + 4/x^6)^(1/6) - 1] / (1/x)
= lim(t→0+) [(1 + 3t + 4t^6)^(1/6) - 1] / t, letting t = 1/x.
Applying L'Hopital's Rule (since this is of the form 0/0) yields
lim(t→0+) [(1 + 3t + 4t^6)^(-5/6) * (3 + 24t^5) - 0] / 1
= lim(t→0+) (3 + 24t^5)/(1 + 3t + 4t^6)^(5/6)
= 3/1
= 3.
I hope this helps!
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Ah yes; there should be a factor of 1/6 from the power rule.
This factor will make the answer 1/2...
This factor will make the answer 1/2...
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