91Ó°ÊÓ

First, we will write the sequence in limit notation. We want to find the limit as n approaches infinity: $$\lim_{n\to\infty}\left(1+\frac{4}{n}\right)^{3 n}$$

We can rewrite the expression inside the parenthesis as the exponent of the natural logarithm (ln) to make it easier to work with: $$\lim_{n\to\infty}e^{3n\ln\left(1+\frac{4}{n}\right)}$$ Now define a new function, y: $$y=3n\ln\left( 1+\frac{4}{n} \right)$$ Take the natural logarithm of both sides: $$\ln y=\ln\left(e^{3n\ln\left(1+\frac{4}{n}\right)}\right) = 3n\ln\left(1+\frac{4}{n}\right)$$ Now, we will differentiate both sides with respect to n and then divide: $$\frac{d}{dn}\left(\ln y\right) = \frac{1}{y}\frac{dy}{dn} = \frac{d}{dn}\left(3n\ln\left(1+\frac{4}{n}\right)\right)$$ Differentiate the right-hand side: $$\frac{1}{y}\frac{dy}{dn} = 3\ln\left(1+\frac{4}{n}\right) - \frac{12}{n\left(1+\frac{4}{n}\right)}$$ Now, plug in the limit as n approaches infinity. Since y = e, we will ignore the left-hand side and focus on simplifying the right-hand side: $$\lim_{n\to\infty} 3\ln\left(1+\frac{4}{n}\right) - \frac{12}{n\left(1+\frac{4}{n}\right)}$$ Using L'Hopital's rule, we get: $$\lim_{n\to\infty} 0 - \frac{0}{\infty} = 0$$ Now, to find y, exponentiate both sides: $$y=e^0 = 1$$

Finally, substitute back the solution (y = 1) into the original limit expression: $$\lim_{n\to\infty}\left(1+\frac{4}{n}\right)^{3 n} = 1$$ So, the limit of the sequence is 1.