# Limit of $\underset{n\to \infty }{\text{lim}}\frac{\ln (n+1)}{\ln (n)}$ without L'Hôpital [duplicate]

I intuitively understand that the limit goes to 1 and I can solve with L'Hôpital but I can't without it.

I tried to call it equal to y and rise both sides to the base e but doesn't seems to work.

$$\underset{n\to \infty }{\text{lim}}\frac{\ln (n+1)}{\ln (n)}$$

Hint: $$\displaystyle\log(n+1)=\log\left(n\left(1+\frac1n\right)\right)=\log(n)+\log\left(1+\frac1n\right)$$
You can use Taylor's formula $$\dfrac{\ln(1+n)}{\ln(n)}=1+\dfrac{1}{\ln n}\ln(1+\frac{1}{n})=1+\dfrac{1}{n\ln n}+\dfrac{1}{n^2\ln n}O(1)$$ as $$n\to \infty$$
The way I thought of doing it is subtracting the expression by $$1$$ and then adding it back at the end. After we subtract $$1$$, the numerator becomes $$\ln (n+1) - \ln (n)$$, which when we use the logarithm subtraction rule, we get $$\ln (\frac{n+1}{n})$$. Since $$\lim_{x\to\infty}$$ of $$\frac{n+1}{n}$$ is $$1$$, the numerator approaches $$0$$, while the denominator which is $$\ln(n)$$ approaches $$\infty$$. Therefore, after subtracting $$1$$, the limit becomes $$0$$, so the value of the limit is $$1$$.
$$\frac{\ln (n+1)}{\ln n} = \frac{\ln n(n+1/n)}{\ln n} = 1+ \frac{\ln (1+1/n)}{\ln n}$$ and $$\lim\limits_{n \to \infty}\ln (1+1/n) = 0$$.