91Ó°ÊÓ

We will use the properties of logarithms to rewrite the given expression. Specifically, we will use the following property: $$\ln(a) + \ln(b) = \ln(ab)$$ So we can rewrite the sequence as: $$\\{ \ln{\left(\sin(\frac{1}{n})\cdot n\right)} \\}$$

Now our sequence looks like this: $$\\{ \ln{\left(n\sin(\frac{1}{n})\right)} \\}$$ Notice that we have the function inside the logarithm: $$f(n)=n\sin(\frac{1}{n})$$ Our goal in this step is to find the limit of this function as n approaches infinity, and then find the limit of the logarithm.

Let's find the limit of f(n) as n approaches infinity using the squeeze theorem. The squeeze theorem states that if for all numbers in an interval, we have $$g(n) \leq f(n) \leq h(n)$$ and $$\lim_{n \to \infty}g(n) = \lim_{n \to \infty} h(n) = L$$, then $$\lim_{n \to \infty} f(n) = L$$. We will find two functions g(n) and h(n) such that f(n) is between them. We notice that: $$-1 \leq \sin(\frac{1}{n}) \leq 1$$ Multiplying all 3 sides by n (since n is positive), we have: $$-n \leq n\sin(\frac{1}{n}) \leq n$$ So, we can compare f(n) with g(n) = -n and h(n) = n. Now we have: $$\lim_{n \to \infty} -n = -\infty$$ $$\lim_{n \to \infty} n = \infty$$ Since the limits of g(n) and h(n) are not the same, the squeeze theorem does not provide any result for our function f(n).

Since the squeeze theorem does not help, we will try substitution. Let: $$x = \frac{1}{n}$$ As n approaches infinity, x approaches 0. So, we can rewrite our original sequence as: $$\\{ \ln{\left(\frac{\sin(x)}{x}\right)} \\}$$ Now, find the limit as x approaches 0. This is a well-known limit, and its value is: $$\lim_{x \to 0} \frac{\sin(x)}{x} = 1$$ Therefore, we can rewrite our sequence as: $$\\{ \ln{1} \\}$$

Now we just need to find the limit of the logarithm as n approaches infinity. Since the value inside the logarithm is constant, the limit does not depend on n: $$\lim_{n \to \infty} \ln{1} = \ln{1} = 0$$ Thus, the limit of the given sequence exists and is equal to 0.