24. { (2e^n + 1) / e^n }

Explanation:

Step1: Simplify the expression

We can rewrite $\frac{2e^{n}+1}{e^{n}}$ as $\frac{2e^{n}}{e^{n}}+\frac{1}{e^{n}}$. Since $\frac{2e^{n}}{e^{n}} = 2$ and $\frac{1}{e^{n}}=e^{-n}$, the sequence becomes $a_{n}=2 + e^{-n}$.

Step2: Find the limit as $n\to\infty$

We know that $\lim_{n\to\infty}e^{-n}=\lim_{n\to\infty}\frac{1}{e^{n}} = 0$ (because the exponential - function $y = e^{n}$ grows without bound as $n\to\infty$). Then $\lim_{n\to\infty}(2 + e^{-n})=\lim_{n\to\infty}2+\lim_{n\to\infty}e^{-n}$. Since $\lim_{n\to\infty}2 = 2$ and $\lim_{n\to\infty}e^{-n}=0$, we have $\lim_{n\to\infty}(2 + e^{-n})=2$.

Answer:

The limit of the sequence $\left{\frac{2e^{n}+1}{e^{n}}\right}$ as $n\to\infty$ is 2.