91Ó°ÊÓ

Observe the numerator, which is given by $$e^{-n}$$. As n approaches infinity, $$e^{-n}$$ approaches 0. This is because the exponential function decays to zero as the exponent becomes increasingly negative (also the limit definition of the exponential decay).

Looking at the denominator, which is given by $$2 \sin \left(e^{-n}\right)$$. Notice, as n approaches infinity, the argument inside the sine function also approaches 0, since by a conclusion from step 1, we know $$e^{-n}$$ goes to 0 as n approaches infinity. Therefore, the limit of the denominator will be $$2 \sin(0)$$, which equals to 0.

When trying to find the limit of a sequence as a fraction, it is important to analyze the behavior of the numerator and denominator. Since both the numerator and denominator approach 0 as n approaches infinity, we need to use L'Hopital's rule, which states that if the limit of the numerator and denominator are both 0 or infinity, then: $$\lim_{n\to\infty}\frac{f(n)}{g(n)}=\lim_{n\to\infty}\frac{f'(n)}{g'(n)}$$, provided the right-hand-side exists. First, we calculate the derivatives of the numerator and denominator with respect to n: $$\frac{d}{dn} \Big(e^{-n}\Big)= -e^{-n}$$ $$\frac{d}{dn} \Big(2 \sin\left(e^{-n}\right)\Big)= -2e^{-n}\cos\left(e^{-n}\right)$$ Now, applying L'Hopital's Rule: $$\lim_{n\to\infty}\frac{e^{-n}}{2 \sin \left(e^{-n}\right)}=\lim_{n\to\infty}\frac{-e^{-n}}{-2e^{-n}\cos\left(e^{-n}\right)}$$ Let's simplify this fraction: $$\lim_{n\to\infty}\frac{1}{2\cos\left(e^{-n}\right)}$$ As n approaches infinity, the argument inside the cosine function approaches 0. Hence, the limit of the denominator will be $$2\cos(0)$$, which equals to 2. Therefore, our limit converges to$$\frac{1}{2}$$. Our limit is: $$\lim_{n\to\infty}a_{n}=\frac{1}{2}$$

Using a graphing utility, plot the sequence $$a_{n}=\frac{e^{-n}}{2 \sin \left(e^{-n}\right)}$$ to check if it converges to our calculated limit. As n approaches infinity, we should see that the sequence converges to the value of $$\frac{1}{2}$$, confirming our result.