$\lim _{x \rightarrow \infty}\left(\frac{x+6}{x+1}\right)^{x+4}=e^{n}$ n=____?

Question

Professor PC Thomas Classes · Answer

We know that $\lim _{x ightarrow \infty}\left(1+\frac{m}{x}ight)^{x}=e^{m}$ Now $\lim _{x ightarrow \infty}\left(\frac{x+6}{x+1}ight)^{x+4}=\lim _{x ightarrow \infty}\left(\frac{1+6 / x}{1+\frac{1}{x}}ight)^{x+4}$\ $=\lim _{x ightarrow \infty}\left(\frac{1+\frac{6}{x}}{1+\frac{1}{x}}ight)^{x}\left(\frac{1+\frac{6}{x}}{1+\frac{1}{x}}ight)^{4}$\ $=\lim _{x ightarrow \infty}\left(\frac{1+\frac{6}{x}}{1+\frac{1}{x}}ight)^{x} 	imes 1$\ $=\frac{\lim _{x ightarrow \infty}\left(1+\frac{6}{x}ight)^{x}}{\lim _{x ightarrow \infty}\left(1+\frac{1}{x}ight)^{x}}=\frac{e^{6}}{e}=e^{5}$\