Question: I have a problem with this limit:

$$\lim\_{x\to 0} \frac{(\ln(1+x\_2)-x\_2)(1-\cos(x\_3))\tan(x\_1)}{\|x\|^4} $$

where $\|\cdot\|$ indicates the Euclidean norm and $x\in\Bbb R^3$.

I have used Taylor expansion and have gone so far to:

$$\lim\_{x\to 0} \frac{(-\frac{x^2\_2}{2}+o(x^2\_2))(\frac{x^2\_3}{2}+o(x^2\_3))(x\_1+\frac{x^3\_1}{3}+o(x^3\_1))}{(x^2\_1+x^2\_2+x^2\_3)^2}$$

It doesn't seem to be hard, but I am a little confused about bottom indexes (which I interpret as multivariables), thus I would ask you for some explanation.

Answer: 
As you note, the numerator is asymptotic to $-\tfrac14x\_1x\_2^2x\_3^2$, which has $\pm O(\Vert x\Vert^5)$ bounds. This gives $\pm O(\Vert x\Vert)$ bounds on the fraction, making the limit $0$ by the squeeze theorem.
We have that

$$\frac{(\ln(1+x\_2)-x\_2)(1-\cos(x\_3))\tan(x\_1)}{\|x\|^4}= \frac{\ln(1+x\_2)-x\_2}{x\_1^2} \frac{1-\cos(x\_3)}{x\_3^2} \frac{\tan(x\_1)}{x\_1} \frac{x\_1^2x\_3^2x\_1}{\|x\|^4}\to 0$$

indeed by standard limits

* $\frac{\ln(1+x\_2)-x\_2}{x\_1^2}\to -\frac12$
* $\frac{1-\cos(x\_3)}{x\_3^2}\to \frac12$
* $\frac{\tan(x\_1)}{x\_1}\to 1$

and by spherical coordinates

$$\frac{x\_1^2x\_3^2x\_1}{\|x\|^4}=\frac{x\_1^2x\_3^2x\_1}{(x\_1^2+x\_2^2+x\_3^2)^2}=\rho \cos^2 \theta\sin^2 \theta\sin^4 \phi \cos \phi \to 0$$