Calculate the limit, Euclidean norm, multivariable function $\lim_{x\to 0} \frac{(\ln(1+x_2)-x_2)(1-\cos(x_3))\tan(x_1)}{\|x\|^4} $

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 1

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.

Answer 2

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

$$\frac{x_1^2x_3^2x_1}{\|x\|^4}=\frac{x_1^2x_3^2x_1}{(x_1^2+x_2^2+x_3^2)^2}=\rho \cos^2 \theta\sin^2 \theta\sin^4 \phi \cos \phi \to 0$$