I'd like to know if my proof is right for this exercise.
Prove using the definitions (without using limits):
$15n+\frac{n}{\log_2n}+\sqrt{n} \in \Theta(n)$
Proof:
We are looking for constants $c_1, c_2, n_0>0$ such that for every $n \ge n_0$ :
$0\lt c_1n\le15n+\frac{n}{\log_2n}+\sqrt{n}\le c_2n$
For the left inequality we'll choose $c_1 = 1$ and it will work for every $n>1$.
For the right inequality:
$15n+\frac{n}{\log_2n}+\sqrt{n}\le c_2n \iff 15+\frac{1}{\log_2n}+\frac{1}{\sqrt{n}}\le c_2 \iff \frac{1}{\log_2n}+\frac{1}{\sqrt{n}}\le c_2 -15$
We'll choose $c_2=16$ and get:
$\frac{1}{\log_2n}+\frac{1}{\sqrt{n}}\le 1$
Which applies for every $n\ge4$.
Therefore, we'll choose $n_0 = 4$ and both inequalities are true for every $n\ge n_0$.
$\blacksquare$