How can one show that $\displaystyle{\lim_{n \to \infty}\left(2 + {1 \over n}\right)^n}$ ?.
I am trying to shoehorn the definition of $e$ somewhere but fruitlessly so.
How can one show that $\displaystyle{\lim_{n \to \infty}\left(2 + {1 \over n}\right)^n}$ ?.
I am trying to shoehorn the definition of $e$ somewhere but fruitlessly so.
$$\displaystyle \lim_{n \to \infty} \left(2 + \frac 1n\right)^n\ge\lim_{n \to \infty} 2^n=+\infty$$
$$\left(2+\frac{1}{n}\right)^n = 2^n \left(1 + \frac{1/2}{n}\right)^n \sim 2^n \sqrt{e}$$
Since $\frac{1}{n}$ will asymptotically approach 0, the limit inside the parenthesis will go to 2. In turn, the limit of $2^n$ as $n$ approaches infinity is infinity as that function increases without bound.
Applying L'Hopital $\lim(2+\frac{1}{n})^n=\lim(e^{\log(2+\frac{1}{n})n})=e^{\lim(\log(2+\frac{1}{n})/n^{-1})}$ the exponent will blow up.