Seeking a rigorous proof for a limit

Question

Prove that$$\lim_{n\to\infty}\frac{\ln (1- \frac{3}{n})}{n}=0$$

I knew that this is the indeterminate form $0/\infty$(Actually it isn't, so I made a dumb mistake) that should be zero but was unable to prove it. I haven't tried using the definition yet because I feel that's too cumbersome?

Answer 1

Note that the limit is not an indeterminate form, indeed:

$$\ln \left(1- \frac{3}{n}\right)\to \ln 1=0 \implies \frac{\ln (1- \frac{3}{n})}{n}\to\frac{0}{+\infty}=0$$

Answer 2

For $n\ge 4$, since the sequence $\{n/(n-3)\}$ is decreasing and the function $\ln$ is increasing, $$\left|\ln\left(1-\frac3n\right)\right|=\left|\ln\left(\frac n{n-3}\right)\right|\le\ln 4$$

Given $\epsilon>0$, take $n_0 >\max(4,\frac{\ln 4}\epsilon)$. If $n\ge n_0$ then $$\left|\frac{\ln\left(1-\cfrac3n\right)}n\right|\le\frac{\ln 4}{n_0}<\epsilon$$

Answer 3

Ah, guys, I have overcomplicated the issue. I just found an easy way to do the proof.

Because $\lim_{x\rightarrow\infty}\ln{(1-\frac{3}{n})=0}$, $\forall \epsilon\in\mathbb R$, $\exists n_o$ such that $\forall n>n_0$, $-\epsilon<\ln{(1-\frac{3}{n})}<\epsilon$. Thus, $\frac{-\epsilon}{n}<\frac{\ln{(1-\frac{3}{n})}}{n}<\frac{\epsilon}{n}$. By the squeeze theorem we easily conclude that the limit tends to 0.

Answer 4

Since$$\lim_{n\to\infty}\frac{\ln\left(1-\frac3n\right)}n=\lim_{n\to\infty}\ln\sqrt[n]{1-\frac3n}$$and since$$(\forall n\in\mathbb{N}\setminus\{1,2,3\}):\frac14\leqslant1-\frac3n<1,$$we have$$\lim_{n\to\infty}\sqrt[n]{1-\frac3n}=1$$and therefore$$\lim_{n\to\infty}\ln\sqrt[n]{1-\frac3n}=0.$$

Answer 5

$I := \displaystyle \int_{1}^{(1-3/n)}(\dfrac{1}{x})dx = \ln(1-3/n)$.

$ |(1)(3/n)| \lt |I| \lt |(3/n)(1-3/n)|$.

Hence :

$(1/n)3/n \lt (1/n)|\ln(1-3/n)| \lt $

$(1/n)(3/n)(1-3/n)$, or

$\dfrac{3}{n^2} \lt \dfrac{\ln(1-3/n)}{n} \lt \dfrac{3(1-3/n)}{n^2}.$

The limit $n \rightarrow \infty$ is ?

Answer 6

Although the limit is not an indeterminate form, as an alternative, by the standard limit

$$x\to 0 \implies \frac{\log(1+x)}{x}\to1$$

we have

$$\frac{\ln (1- \frac{3}{n})}{n}=-\frac{3}{n^2}\frac{\ln (1- \frac{3}{n})}{-\frac3n}\to0\cdot1=0$$