how can I calculate: $\lim_{n \to \infty} \frac{\frac{1}{n}-\ln(1+\frac{1}{n})}{n^{\frac{1}{n}}-1} $?

Question

how can I calculate:
$$\lim_{n \to \infty} \frac{\frac{1}{n}-\ln(1+\frac{1}{n})}{n^{\frac{1}{n}}-1} $$

I tried with Hospital and it's not working. Can help please ?

Answer 1

Hint: put $n = \frac1t$ denominator will go towards $-1$ and numerator will become $0$ giving answer as $0$

Answer 2

I think the L'Hospital helps: $$\begin{align}\lim_{n\rightarrow+\infty}\frac{\frac{1}{n}-\ln\left(1+\frac{1}{n}\right)}{n^{\frac{1}{n}}-1}&=\lim_{x\rightarrow0^+}\frac{x-\ln(1+x)}{\left(\frac{1}{x}\right)^x-1}=\lim_{x\rightarrow0^+}\frac{x-\ln(1+x)}{1-x^x}\\&=-\lim_{x\rightarrow0^+}\frac{1-\frac{1}{1+x}}{x^x(1+\ln{x})}=-\frac{1}{2}\lim_{x\rightarrow0^+}\frac{x}{1+\ln{x}}=0\end{align}$$

Answer 3

Using Taylor expansion

We have $$\frac{1}{n}-\ln\left(1+\frac{1}{n}\right)= \frac{1}{2n^2} +o\left(\frac{1}{n^2}\right)$$ since $$\ln(x+1) = x-\frac{x^2}{2} +o(x^2)$$

and similarly we have, $$n^{1/n} = \exp(n\ln (1/n)) =\exp\left(n\left(\frac{1}{n}-\frac{1}{2n^2}+o\left(\frac{1}{n^2}\right)\right)\right) = 1 -\frac{1}{2n} +o\left(\frac{1}{n}\right)$$

hence, $$ \frac{\frac{1}{n}-\ln(1+\frac{1}{n})}{n^{\frac{1}{n}}-1}= \frac{\frac{1}{2n^2} +o\left(\frac{1}{n^2}\right)}{-\frac{1}{2n} +o\left(\frac{1}{n}\right)} =-\frac{1}{n}+o\left(\frac{1}{n}\right) \to 0~~ as~~~ n\to\infty$$

$$\lim_{n \to \infty} \frac{\frac{1}{n}-\ln(1+\frac{1}{n})}{n^{\frac{1}{n}}-1} =0$$

Answer 4

We will assume that we are equipped with the "standard limits"

$$\color{blue}{\lim_{x\to 0}\frac{e^x-1}{x}=1} \tag 1$$

and

$$\color{red}{\lim_{x\to 0}\frac{\log(1+x)}{x}=1}\tag2$$

Using $(1)$ and $(2)$ with $x=1/n$ and $n\to \infty$, we find that

$$\begin{align} \lim_{n\to \infty}\frac{\frac1n -\log\left(1+\frac1n\right)}{n^{1/n}-1}&=\lim_{n\to \infty}\frac{\frac1n -\log\left(1+\frac1n\right)}{e^{\frac1n \log(n)}-1}\\\\ &=\lim_{n\to \infty}\frac{1-\color{red}{\frac{\log\left(1+\frac1n\right)}{\frac1n}}}{\color{blue}{\frac{e^{\frac1n \log(n)}-1}{\frac1n}}}\\\\ &=\frac{1-1}{1}\\\\ &=0 \end{align}$$

Answer 5

Note that

$$\frac1n-\frac{1}{2n^2}\le\ln(1+\frac{1}{n})\le\frac1n$$

$$n^{\frac{1}{n}}=e^{\frac{\log n}{n}}\ge1+\frac{\log n}{n}$$

thus

$$0=\frac{\frac{1}{n}-\frac1n}{n^{\frac{1}{n}}-1}\le\frac{\frac{1}{n}-\ln(1+\frac{1}{n})}{n^{\frac{1}{n}}-1}\le\frac{\frac{1}{2n^2}}{\frac{\log n}{n}}=\frac{1}{2n\log n}\to 0$$

therefore for squeeze theorem

$$\lim_{n \to \infty} \frac{\frac{1}{n}-\ln(1+\frac{1}{n})}{n^{\frac{1}{n}}-1}=0$$