Limiting Logarithms

You can use the substitution: $\log_{x} y=\frac{\ln{y}}{\ln{x}}$

$\lim_{n\rightarrow \infty} \log_{n}{(n+1) }=\lim_{n\rightarrow \infty} \frac{\ln (n+1)}{\ln{(n)}}$

By L'Hopital's Rule: $\lim_{n\rightarrow \infty} \frac{\ln (n+1)}{\ln{(n)}}=\lim_{n\rightarrow \infty}\frac{\frac{1}{n+1}}{\frac{1}{n}}=\lim_{n\rightarrow \infty} \frac{n}{n+1}=\lim_{n\rightarrow \infty} 1-\frac{1}{n+1}=1$

For $\lim_{x\to \infty}(\log_{n}(n+1))$ Use L'Hopital's theorem for $\lim_{x\to a}\left(\frac{f(x)}{g(x)}\right)$ , if $\lim_{x\to a}\left(\frac{f(x)}{g(x)}\right) =\frac{0}{0}$ or $\lim_{x\to a}\left(\frac{f(x)}{g(x)}\right) =\pm \frac{\infty}{\infty}$ then $\lim_{x\to a}\left(\frac{f(x)}{g(x)}\right) =\lim_{x\to a} \left(\frac{f'(x)}{g'(x)}\right)$

For extremely large n, $\log(n+1) \approx \log(n)$

Thus answer is unity.