Solving a limit using only special limits and algebric manipulations I'm wondering how to solve this limit:
$$\lim_{x \to 0^+} \frac{\tan^3((1+x^{\frac 23})^\frac13-1)+\ln(1+\sin^2(x))}{\arctan^2(3x)+5^{x^4}-1}(\sqrt{\frac{1+x+x^2}{x^2}}-\frac 1x)$$
With my actual notions that are:
-Special limits
-A limit of a sum/product/quotient of functions is the sum/product/quotient of limits of those functions if the functions converge(and also if the denominator function doesn't converge to 0 in the case of quotient)
-Basic notions like $+\infty\cdot a=+\infty, a>0$ etc
-Comparison theorem
-Algebric manipulations
Often my teacher does this "trick":
"If we have to calculate: $\lim_\limits{x \to x_0} s(x)c(x)$. Where $s$ is a simple function that we know to be convergent to a non-zero value and $c$ is a complicated functions whom limit is unknown. We can write this:
$$ \lim_\limits{x \to x_0} s(x)c(x)=\lim_\limits{x \to x_0} s(x)\lim_\limits{x \to x_0} c(x)$$ If we discover then that: $$\lim_\limits{x \to x_0} c(x)\in \mathbb{R}$$
Then our previous passage is justified. If we discover that:
$$\lim_\limits{x \to x_0} c(x)\in \pm \infty$$
Then our previous passage is not justified formally, but it doesn't affect the limit(it's a kind of notation abuse). If we discover that:
$$\not\exists \lim_\limits{x \to x_0} c(x)$$
Then our passage is not justified and it may have affected the limit result"
I kinda understood why this works(it's a kind of retrospective justificatin) but i was wondering if there a was a more formal way to describe this, because when i try to do limits I always try to justify all the steps I do and to be formal. However let's go back to the initial limit and to my attempt:
$$\lim_{x \to 0^+} \frac{\tan^3((1+x^{\frac 23})^\frac13-1)+\ln(1+\sin^2(x))}{\arctan^2(3x)+5^{x^4}-1}(\sqrt{\frac{1+x+x^2}{x^2}}-\frac 1x)$$
Let's try to calculate first:
$$\lim_{x \to 0^+} \sqrt{\frac{1+x+x^2}{x^2}}-\frac 1x=\lim_{x \to 0^+} \frac{\sqrt{1+x+x^2}-1}{x}=\lim_{x \to 0^+} \frac{\sqrt{1+x+x^2}-1}{x+x^2}(x+1)$$
Now i use a known special limit:
$$\lim_{x \to 0^+} \frac{x+1}{2}=\frac 12$$
Now let's use the trick of my teacher and let's hope that the remaining limit exists otherwise we are at the starting point(this is also why sometimes i'm a bit unsure doing this it feels like a bet):
$$\frac 12\lim_{x \to 0^+} \frac{\tan^3((1+x^{\frac 23})^\frac13-1)+\ln(1+\sin^2(x))}{\arctan^2(3x)+5^{x^4}-1}$$
And now i'm stuck because I see many useful special limits that i could apply but it always come to a $$0 \cdot \infty$$ form where i can't apply the "trick". Sometimes I feel i'm overcomplicating everything by being too formal but I really want to understand why I can apply something and I don't want to make it become an automatism before i totally understood it.
 A: This is a typical example which is designed to intimidate students.
You have already noted that the last factor tends to $1/2$. Without assuming anything about rest of the expression you can move this factor out of the limit to get $$\frac{1}{2}\lim_{x\to 0^{+}} \text{ (rest of the expression)} $$ The next part is to simplify the denominator. Let's write $$\arctan^23x+5^{x^4}-1=(9x^2)\left(\left(\frac{\arctan 3x}{3x}\right)^2+\frac{5^{x^4}-1}{x^4}\cdot\frac{x^2}{9}\right)$$ The expression in large parentheses tends to $$1^2+(\log 5)\cdot 0=1$$ and thus this factor can be safely replaced by $1$ and the denominator simplifies to $9x^2$.
Since the numerator consists of two terms we can now split the expression into two parts the simpler of which is $$\frac{\log(1+\sin^2x)}{9x^2}=\frac{1}{9}\cdot\left(\frac{\sin x} {x} \right) ^2\cdot \frac{\log(1+\sin^2x)}{\sin^2x}$$ and this tends to $(1/9)\cdot 1^2\cdot 1=1/9$. Thus your desired limit equals $$\frac{1}{18}+\frac{1}{18}\lim_{x\to 0^{+}}\frac{\tan^3((1+x^{2/3})^{1/3}-1)}{x^2}$$ The expression under limit above can be written as $$\left(\frac{\tan((1+x^{2/3})^{1/3}-1)}{(1+x^{2/3})^{1/3}-1}\right)^3\cdot\left(\frac{(1+x^{2/3})^{1/3}-1}{(1+x^{2/3})-1}\right)^3$$ which tends to $1^3(1/3)^3=1/27$. Thus the desired limit is $$\frac{1}{18}+\frac{1}{18}\cdot\frac{1}{27}=\frac{14}{243}$$

The trick of your teacher works and has been discussed by me in this post. Apart from this you need the rule for limit of composition of functions.

Theorem: If $$\lim_{x\to a} g(x) =b, \lim_{x\to b} f(x) =L$$ and $g(x) \neq b$ as $x\to a$ then $$\lim_{x\to a} f(g(x)) =L$$

