I have a task to show by definition that
$\ln n^{2} + 1 = \Theta (\ln n)$
by definition , we know that we have to find c1 and c2 such as
$c_{1} * \ln n<\ln n^{2} + 1 < c_{2} * (\ln n)$
we can find find c2 easily
$\ln n^{2} + 1 < c_{2} * (\ln n)$
$2*\ln n + 1 < c_{2} * (\ln n)$
so we know that c2 hast to be c2 >= 3.
c1 is the same story
$c_{1} * \ln n< 2*\ln n + 1 $
we see that c1 <= 2
My question is from students point of view. Is it legit to assume that c2 >= 3 and 0 < c1 <= 2 because we "see it" or do we have to prove it by some further calculation?
As
$$\lim_{n\to +\infty}\frac{ln(n^2)+1}{ln(n)}=$$
$$\lim_{n\to+\infty}(2+\frac{1}{ln(n)})=\color{green}{2}$$
with$\;\;\epsilon=1$ for example,
$$\exists N>0: \forall n\geq N$$
$$1<\frac{ln(n^2)+1}{ln(n)}<3$$
So, you can take $c_1=1\;$ and$\;c_2=3$ .