Is this operation of limits wrong? I was taught that the limit addition property $\lim_{x\to a} f(x) + \lim_{x \to a} g(x) = \lim_{x \to a} \left( f(x)+g(x) \right) $ holds only when these limits by themselves aren't infinities with opposite signs, in which case I would end up with $\infty - \infty$, which doesn't make sense. I've found the following in the book Inside interesting integrals, section 1.6, it has a sum of two limits but this is $\infty - \infty$ because of $\lim_{x \to 0} \ln{x}$ so we shouldn't be able to apply the mentioned addition property, but I assume this is what the author did, as he added both logarithms together. Check here (used antiderivative and then split into two integrals. This shouldn't be allowed either because it's expressing a convergent integral as difference of divergent integrals):

He gets the correct result eventually but that doesn't necessarily mean the process is clean. I think this is wrong but I decided to ask since it would be strange to find something wrong like that in a book, so I wanted to make sure. (If this is actually correct please explain why)
Edit: Thanks to user, I was wrong when assuming the original integral was convergent. It turns out it's divergent by standard integration.
 A: There are two things, Limit of the sum and sum of the limits. They are equal only both the limits exist finitely or when both of them diverge to infinity with the same sign. In the book, one diverges to positive infinite and the other one to negative infinite. So, sum of limits$\neq$ limit of sum.
The true limit is always the limit of the sum. So, adding the functions first to get a single function and applying the limit is valid.
The author has first added them into a single function and then applied the limits. So, that is a valid operation.
Similarly, as answered by @user, the limit $$(\lim_{x\to ∞}(1+sinx))+(\lim_{x\to ∞}(1-sinx))\neq\lim_{x\to ∞}2$$
The true limit is always the Limit of the sum, not the sum of limits.
A: Indeed it is wrong, let consider for example
$$f(x)=1+\sin x \quad g(x)=1-\sin x$$
what is true is that if $\lim_{x\to a} f(x)$ and $\lim_{x \to a} g(x) $ both exist finite or to infinite with the same sign then the sum is equal to $\lim_{x \to a} \left( f(x)+g(x) \right) $, extending the sum also for the $\infty$ case with the same sign.
With reference to the example presented, according to the standard definition, the two integrals should be evaluated separetely and since each one diverges also the integral as a whole diverges. Refer also to the convergence definition given here.
