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Veseljchak [2.6K]
3 years ago
7

Determine if the sequence converges: [ln(n)]^2 /n

Mathematics
1 answer:
Finger [1]3 years ago
8 0
We will use l´Hopital´s rule for calculating limits involving indeterminate form (in this case: ∞ / ∞ ) using the derivative of the numerator and denominator:\lim_{n \to \infty}  \frac{ln^{2}n }{n}  =  \lim_{n \to \infty} \frac{2ln(n)* \frac{1}{n} }{1} =  \lim_{n \to \infty}  \frac{2ln(n)}{n} This is still form ∞/∞ and we will use the derivative again:\lim_{n \to \infty}  \frac{2ln(n)}{n}  =  \lim_{n \to \infty}  \frac{ \frac{1}{n} }{1} = \lim_{n \to \infty}  \frac{1}{n}=1/∞ = 0
The sequence converges.
 
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Answer:

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Step-by-step explanation:

Given formula:

V(t) = 420,000 * (0.965)^t

A. What would be worth the car′s worth in 2 years?

V(t) = 420,000 * (0.965)^t

We replace t by 2, as follows:

V(2) = 420,000 * (0.965)²

V(2) = 420,000 * 0.931225

<u>V (2) = $ 391,114.50</u>

B. In how many years will the car be worth $325,000?

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We replace V(t) by 325,000, as follows:

325,000 = 420,000 * (0.965)^t

325,000/420,000 = (0.965)^t

0.77381 =  (0.965)^t

t = log 0.965(0.7738)  

t = log 0.7738/log 0.965  

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