Correctly prove that $\log(+1) = \Theta(\log())$

Question

What is the correct mathematical reasoning to omit the $+ 1$ in $\log(+1) = \Theta(\log())$ since it is a constant?

Answer 1

If $f=\Theta(g)$, then by the limit definition

$$0 < \liminf_{x \to \infty} \left|\frac{f(x)}{g(x)}\right| \le \limsup_{x \to \infty} \left|\frac{f(x)}{g(x)}\right|< \infty.$$

Note that if we choose $f=\log(n+1)$ and $g=\log(n)$, $$\lim_{n\to\infty}\left|\frac{\log(n+1)}{\log(n)}\right| = \liminf_{n \to \infty} \left|\frac{\log(n+1)}{\log(n)}\right| = \limsup_{n \to \infty} \left|\frac{\log(n+1)}{\log(n)}\right|.$$

The $+1$ does not change the growth of the function, so $\log(n+1)/\log(n)=1$ as $n\to\infty.$

In conclusion we have $$0<1\le1<\infty,$$ which is true and so $\log(n+1)=\Theta(\log n)$.

Answer 2

As $n+1<n^2$ for $n>2$, we have $\ln(n+1)<2\ln n$ and so $\ln(n+1)=O(\ln n)$. On the other hand, $\ln n=O(\ln(n+1))$ follows from $\ln n<\ln(n+1)$.