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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.

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

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Hint: $\displaystyle\log(n+1)=\log\left(n\left(1+\frac1n\right)\right)=\log(n)+\log\left(1+\frac1n\right)$

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    $\begingroup$ That's exactly what I was looking for. Many thanks! $\endgroup$ Apr 1 at 19:16
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    $\begingroup$ I'm glad I could help. $\endgroup$ Apr 1 at 19:44
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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 $n\to \infty$

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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 $1$.

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You have

$$\frac{\ln (n+1)}{\ln n} = \frac{\ln n(n+1/n)}{\ln n} = 1+ \frac{\ln (1+1/n)}{\ln n}$$ and $\lim\limits_{n \to \infty}\ln (1+1/n) = 0$.

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