Final Answer:
The limit of the given sequence as n approaches infinity is -10.
Step-by-step explanation:
To find the limit of the given expression, we need to evaluate the limit of the logarithmic function as n approaches infinity. We can start by analyzing the expression inside the logarithm.
The first term, ln(n3+3), can be written as:
ln(n3+3) = ln(n3) + ln(3)
The limit of ln(n3) as n approaches infinity is 0, since the limit of n3 as n approaches infinity is infinity, and the logarithm function approaches infinity as its argument approaches infinity.
The limit of ln(3) as n approaches infinity is 1, since 3 is a constant.
Therefore, the limit of the first term is:
lim(n → ∞) ln(n3+3) = lim(n → ∞) (ln(n3) + ln(3)) = 0 + 1 = 1
Now, let’s analyze the second term, ln(2n3+11n). We can write it as:
ln(2n3+11n) = ln(2n3) + ln(11n)
The limit of ln(2n3) as n approaches infinity is 0, since the limit of n3 as n approaches infinity is infinity, and the logarithm function approaches infinity as its argument approaches infinity.
The limit of ln(11n) as n approaches infinity is 11, since 11 is a constant.
Therefore, the limit of the second term is:
lim(n → ∞) ln(2n3+11n) = lim(n → ∞) (ln(2n3) + ln(11n)) = 0 + 11 = 11
Finally, we can find the limit of the given expression by subtracting the limit of the second term from the limit of the first term:
lim(n → ∞) (ln(n3+3)−ln(2n3+11n)) = lim(n → ∞) (1 - 11) = -10
Therefore, the limit of the given sequence as n approaches infinity is -10.