Proof of the limit for the sequence of function.

Question

Here I need to prove / disprove that the limit for the sequence of functions is one. Here is what I have found: $$\lim_{n \to \infty} \sqrt[n]{\left(1+\frac{1}{n}\right)^n -e}=1 \\ \text{sequence } \left(1+\frac{1}{n}\right)^n \text{ upper limit is } e \Rightarrow \lim_{n \to \infty} sup = 0 \neq 1 $$

But why is the sup taken in order to disprove that the limit is not equal to 1?

Answer 1

Note that $\left(1+\frac1n\right)^n<e$. Hence, $\left(1+\frac1n\right)^n-e<0$. Inasmuch as $\sqrt[n]{x}$ is not defined for $x<0$ when $n$ is an even integer, the limit is not well defined.

Instead, let's analyze the limit $$\lim_{n\to\infty }\sqrt[n]{e-\left(1+\frac1n\right)^n}$$To proceed, we write

$$\begin{align} e-\left(1+\frac1n\right)^n&=e\left(1-e^{n\log\left(1+\frac1n\right)-1}\right)\\\\ &=e\left(1-e^{-\frac1{2n}+O\left(\frac1{n^2}\right)}\right)\\\\ &=e\left(\frac1{2n}+O\left(\frac1{n^2}\right)\right) \end{align}$$

whereupon taking the $n$th root and letting $n\to \infty$, we find that

$$\lim_{n\to\infty }\sqrt[n]{e-\left(1+\frac1n\right)^n}=1$$

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NOTE: If we are permitted to broaden the definition of the $n$th root to be defined in terms of the complex logarithm, then for any fixed branch of the complex logarithm, the value of the original limit is also $1$.