91Ó°ÊÓ

As n approaches infinity, \(e^{n}\) also approaches infinity. But since the sine function has a periodic nature with a range of \([-1, 1]\), the limit value of the sine function does not exist as n approaches infinity. Therefore, the limit of \(\sin e^{n}\) is \(\lim_{n \rightarrow \infty}\sin e^{n}\).

As n approaches infinity, \(n^2\) also approaches infinity. As a result, the exponential function \(e^{-n^{2}}\) approaches 0. Therefore, the limit of \(e^{-n^{2}}\) is \(\lim_{n \rightarrow \infty}e^{-n^2} = 0\).

To analyze the limit of the fraction, we first notice that both the numerator and the denominator are growing to infinity as n approaches infinity. Therefore, we can apply L'Hopital's Rule to find \(\lim_{n \rightarrow \infty} \frac{n^{2}}{n^{2}+1}\). Taking the derivative of both the numerator and the denominator with respect to n, we get: Numerator: \(\frac{d}{dn}(n^2) = 2n\) Denominator: \(\frac{d}{dn}(n^2 + 1) = 2n\) Now, let's apply L'Hopital's Rule: \(\lim_{n \rightarrow \infty} \frac{2n}{2n} = 1\)

Now that we have analyzed the limit of each subexpression, let's combine them and find the overall limit of the expression: \(f(0) =\lim _{n \rightarrow \infty}\left(\left(\sin e^{n}\right) e^{-n^{2}}+\frac{n^{2}}{n^{2}+1}\right)\) First, we know that \(\lim_{n \rightarrow \infty} \sin e^{n}\) does not have a well-defined limit and \(\lim_{n \rightarrow \infty} e^{-n^2} = 0\). So their product: \(\lim_{n \rightarrow \infty} (\sin e^{n}) e^{-n^{2}} = 0\) Next, we substitute the limit of subexpressions back into the expression: \(f(0)=\lim _{n \rightarrow \infty} \left(0 + \frac{n^{2}}{n^{2}+1}\right) = \lim _{n \rightarrow \infty} \frac{n^{2}}{n^{2}+1}\) As we found in Step 3: \(f(0) = 1\) Thus, the limit of the given expression as \(n\) approaches infinity is 1.