Can I take limits under a limit? Is the following method correct? I know you cannot replace an expression with a sub-expression in a limit, but is this the case? I am talking specifically about lines $2$ and $3$. If this is not the correct way, why is my answer correct? How should I approach such limit? Thank you for any help!$$\lim_{u\to \infty}\left(-\frac{u^2\ln\left(u^2\right)}{1+u^2}-\ln\left(\frac1{1+u^2}\right)\right)$$
$$=\lim_{u\to \infty}\left(-\frac{u^2\ln\left(u^2\right)}{u^2\left(1+\frac1{u^2}\right)}+\ln\left(1+u^2\right)\right)$$
$$=\lim_{u\to \infty}\left(-\ln\left(u^2\right)+\ln\left(1+u^2\right)\right)$$
$$=\lim_{u\to \infty}\ln\left(\frac{1+u^2}{u^2}\right)$$
$$=\lim_{u\to \infty}\ln\left(\frac{u^2\left(1+\frac1{u^2}\right)}{u^2}\right)$$
$$=\lim_{u\to \infty}\ln1=0$$
$$$$***Edit: Is there a way to evaluate this limit $\textbf{without}$ the use of  Series Expansion?
$$$$Maybe this might be useful, the original limit was:
$$\lim_{x\to 0}\left(\frac{\ln\left(x^2\right)}{1+x^2}-\ln\left(\frac{x^2}{1+x^2}\right)\right)$$ 
As you can see, I let: $$x=\frac1{u}$$
As I thought it would be nicer to work with $u\to\infty$
 A: To address your initial question: no, your approach isn't correct, for precisely the reason you stated yourself. It just so happens that this wrong approach gives you the right answer.
Anyway, on towards tackling the limit: $$\frac{\ln x^2}{1+x^2} - \ln x^2 + \ln (1+x^2) = \ln x^2 \left(\frac{1}{1+x^2} - 1\right) + \ln (1+x^2)$$
But $\ln(1+x^2) \to \ln 1 = 0$ as $x\to 0$ by continuity of $\log$. So we need only work on $$\frac{x^2\ln x^2}{1+x^2} = \frac{\ln x^2}{1 + x^{-2}}$$
We could potentially use L'Hopital (since it's of the form $\infty/\infty$) on this to get $$\frac{2x^{-1}}{-2x^{-3}} =-x^2 \to 0$$ so $(x^2\ln x^2)/(1+x^2) \to 0$ so the entire expression goes to $0$, since it's the sum of two things that go to $0$. I'm not sure how to avoid L'Hopital at that last step, but I figured I'd post this anyway, especially re: your last comment.
A: As dxiv pointed it out, step 3 does not follow from step 2 (basically you took the limit in the denominator, without making $u$ tends to $-\infty$ in the numerator, which you aren't allowed to).
One way to find your limit without series expansion (so sad) and without L'Hospital rule would be the following:
$$\dfrac{\ln(x^2)}{1+x^2}-\ln\left(\dfrac{x^2}{1+x^2}\right) = -\dfrac{\ln(x^2) x^2}{1+x^2} + \ln(1+x^2)$$
Then $\ln(x^2)x^2 \rightarrow 0$ when $x$ goes to $0$, that's high school stuff (let $y=x^2$, you have to find the limit of $y \ln(y)$ when $y$ goes to $0$). And $1+x^2 \rightarrow 1$. Then your expression goes to $-0/1 + \ln(1)$, which is $0$.
A: What you are essentially doing is to use the following claim:
$
\def\lfrac#1#2{{\large\frac{#1}{#2}}}
$

(Wrong) $\lim_{x \to p} f(g(x),h(x)) = \lim_{x \to p} f(\lim_{y \to p} g(y),h(x))$.

For a simple but instructive counter-example:

$1 = \lim_{x \to 0} \lfrac{\exp(2x)-(1+x)}{x} \ne \lim_{x \to 0} \lfrac{\exp(2x)-(1)}{x} = 2$ even though $\lim_{x \to 0} (1+x) = 1$.

Also, although I always advise using asymptotic expansion, we can 'avoid' using it in our final solution even though we can derive it via asymptotic expansion. Namely we look at the most significant terms in the expansions and pull them out by subtraction or division as appropriate, such that we can easily deal with the rest.
$\lfrac{\ln(x^2)}{1+x^2}-\ln\big(\lfrac{x^2}{1+x^2}\big)
 = \ln(x^2) - \lfrac{x^2 \ln(x^2)}{1+x^2} - \ln(x^2) + \ln(1+x^2)
 = - \lfrac{x^2 \ln(x^2)}{1+x^2} + \ln(1+x^2)$.
As $x \to 0$, we clearly have $\ln(1+x^2) \to 0$ and $x^2 \ln(x^2) = \lfrac{-\ln(1/x^2)}{1/x^2} \to 0$ since $1/x^2 \to \infty$. You can prove these by your favourite methods, two of which I have sketched in this comment.
Another issue in your question that I realize has not been addressed is that it is incorrect to let $x = \lfrac1u$. Think of it this way. You are given $x \to 0$, meaning that $x$ is going towards $0$ in some manner that you have no control over. All you know is that it eventually remains as close to $0$ as you desire, but never equal. You are free to let $u = \lfrac1x$, but then you cannot claim that $u \to \infty$. It may not since $x$ may oscillate about $0$. But you can let $u = \lfrac1{x^2}$ and claim that $u \to \infty$, because it is true and can be proven.
A: I would like to point out that your calculation is not quite as wrong as you might think. The point is that obviously you cannot replace any subexpression in side a limit with its limit, but there is the algebraic limit theorem (see e.g. wikipedia or this question) which allows certain such operations. For the case at hand, it states that if $\lim_{u\to\infty} f(u)$ and $\lim_{u\to\infty} g(u)$ exist and one of them is finite, then
$$ \lim\limits_{u\to\infty} \!\left(f(u)+ g(u)\right)=\lim\limits_{u\to\infty} f(u) + \lim\limits_{u\to\infty} g(u)$$
and
$$ \lim\limits_{u\to\infty} f(u)\cdot g(u)=\lim\limits_{u\to\infty} f(u) \cdot \lim\limits_{u\to\infty} g(u)\,.$$
For your case, we have 
$$\lim\limits_{u\to\infty} \!\left[-\frac{\ln\left(u^2\right)}{1+\frac1{u^2}}+\ln\left(1+u^2\right)\right]\\
=\lim\limits_{u\to\infty} \!\left[\left(-\ln u^2+\left(1+\frac{1}{u^2}\right)\ln\!\left(1+u^2\right)\right)\cdot\underbrace{\left(\frac{1}{1+\frac{1}{u^2}}\right)}_{g(u)}\right]$$
Now $\lim_{u\to\infty}g(u)=1$, so we can drop it by the product rule above. We are left with
$$\lim\limits_{u\to\infty} \!\left[\ln\left(\frac{1+u^2}{u^2}\right)+\underbrace{\frac{\ln\!\left(1+u^2\right)}{u^2}}_{h(u)}\right]$$
Here, $\lim_{u\to\infty}h=0$, and we can use the sum rule above to discard it and finally arrive at your result. 
