Evaluate: $\lim_{n\to \infty} \bigg[n-{n\over e}\bigg(1+{1\over n}\bigg)^n \bigg] $

Question

$$\lim_{n\to \infty} \bigg[n-{n\over e}\bigg(1+{1\over n}\bigg)^n \bigg] $$

My solution: Can we solve this as $$\lim_{n\to \infty}{\bigg(1+{1\over n}\bigg)^n}=e$$ $$\lim_{n\to \infty} {n\over e}\bigg[e-\bigg(1+{1\over n}\bigg)^n \bigg] =0$$ Can we proceed this way? is there anyother way we can approach? Please help . Thanks in advance.

You can do that!!! In general, $\lim u_n v_n \neq 0$ even if $\lim v_n =0$. An easy example is $u_n =n$ and $v_n =1/n$. However, you can use Taylor expansion at order 2 of $e^{n \ln \left(1+\frac{1}{n}\right)}$ to get the limit you're looking for. — mathcounterexamples.net, Feb 11 '23 at 16:20
you can't because you're dealing with an indetrminate form ($0\cdot \infty$) — Sine of the Time, Feb 11 '23 at 16:40
Also: https://math.stackexchange.com/q/3986021/42969, https://math.stackexchange.com/q/3358294/42969, https://math.stackexchange.com/q/3747770/42969 – all found with Approach0 — Martin R, Feb 11 '23 at 16:48

Mhd Bhsn · Answer 1 · 2023-02-11T16:46:06.703

1

You can take advantage of the Taylor series of $(1+ x)^{\frac{1}{x}}$,which you can find here
So $\Big(1+\dfrac{1}{n}\Big)^n \sim e(1-\dfrac{1}{2n})$
This yields $L = \dfrac12$

edited Feb 11 '23 at 16:46

answered Feb 11 '23 at 16:28

Mhd Bhsn

71

Evaluate: $\lim_{n\to \infty} \bigg[n-{n\over e}\bigg(1+{1\over n}\bigg)^n \bigg] $

1 Answers1