<s>[INST] Is $\frac{\lim_{x \to y} a}{\lim_{x \to y} b} \ne \lim_{x\to y}\frac{a}{b}$? So I had this question

$$\lim_{x \to 0} \frac{\sin(ax) + bx}{\sin(bx) + ax}$$

I did it this way:

$$\lim_{x \to 0} \frac{\sin(ax) + bx}{\sin(bx) + ax}$$
$$= \frac{\lim_{x \to 0}\sin(ax) + \lim_{x \to 0}bx}{\lim_{x \to 0}\sin(bx) + \lim_{x \to 0}ax}$$
$$= \frac{\lim_{x \to 0}\sin(ax)}{\lim_{x \to 0}\sin(bx)}$$
$$= \frac{\displaystyle\frac{\lim_{x \to 0}\sin(ax) \cdot ax}{ax}}{\displaystyle\frac{\lim_{x \to 0}\sin(bx) \cdot bx}{bx}}$$
$$=\frac{\lim_{x \to 0} ax}{\lim_{x \to 0} bx}$$
$$=\lim_{x \to 0}\frac{ax}{bx}$$
$$=\lim_{x \to 0}\frac{a}{b} = \frac{a}{b}$$

Which is incorrect.
I did it again and with a different approach :

$$\lim_{x \to 0} \frac{\sin(ax) + bx}{\sin(bx) + ax}$$
$$= \lim_{x \to 0} \frac{\displaystyle\frac{\sin(ax) + bx}{bx}}{\displaystyle\frac{\sin(bx) + ax}{bx}}$$
$$= \lim_{x \to 0} \frac{\displaystyle\frac{\sin(ax)}{bx} +1}{\displaystyle\frac{\sin(bx) + ax}{bx}}$$
$$= \frac{\lim_{x \to 0} \left(\displaystyle\frac{\sin(ax)}{bx} +1\right)}{\lim_{x \to 0} \left( \displaystyle\frac{\sin(bx) + ax}{bx} \right) }$$
$$= \frac{\lim_{x \to 0} \left(\displaystyle\frac{\sin(ax)}{bx}\right) +1}{\displaystyle\frac{a}{b} + 1}$$
$$= \frac{\lim_{x \to 0} \left(\displaystyle\frac{\sin(ax) \cdot ax}{bx \cdot ax}\right) +1}{\displaystyle\frac{a}{b} + 1}$$
$$= \frac{\displaystyle\frac{a}{b} +1}{\displaystyle\frac{a}{b} + 1} = 1$$

Which is the correct answer.
Can anyone tell me where was my first approach incorrect ?
I am guessing that this line

$$\frac{\lim_{x \to 0} ax}{\lim_{x \to 0} bx} = \lim_{x \to 0}\frac{ax}{bx}$$

is incorrect.
Thanks in advance.
 [/INST] In short, you cannot naively "split" limits. A first reason for that is that they don't necessarily exist. For example, the sequence $0$ has a limit in infinity (here, 0) and $(-1)^n$ has no limit when $n\to\infty$, but $0=(-1)^n-(-1)^n$.
Also, both the numerator and denominator may have a limit, but what if the denominator's limit is 0? Then the ratio is not defined.
An efficient way to deal with such limits is to use Taylor's expansion. Here,
$\sin(ax)+bx=ax-(ax)^3/6+bx+o(x^3)$ and $\sin(bx)+ax=bx-(bx)^3+ax+o(x^3)$, so, (assuming $a\neq -b$, otherwise, it's easy):
\begin{align}\dfrac{\sin(ax)+bx}{\sin(bx)+ax}&=\dfrac{ax+bx-a^3x^3/6}{bx+ax-b^3x^3/6}+o(x^2)=\dfrac{\frac{1}{b+a}(a+b-a^3x^2/6)}{\frac{1}{b+a}(b+a-b^3x^2/6)}+o(x^2) \\ &=1+\dfrac{b^3-a^3}{6(a+b)}x^2+o(x^2) & \\\end{align}
This proves that the limit in $0$ is $1$, and also that the way the fraction approaches 1 is parabolic is the neighbourhood of 0.
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