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I'm wondering if my approach to solving this problem is correct and not violating any rules. So in an exercise I was asked to check if the limit:
$$\lim_{\bar x\to \bar 0} \frac{\ln(1+\vert \bar x \vert^2)}{\vert \bar x \vert^2 + \sin(x_1x_2x_3)}$$ exists and if so what it is, where $\bar x = (x_1,x_2,x_3)$ and $\vert \bar x \vert = \sqrt {x_1^2 + x_2^2 + x_3^2} $

Instead of checking different approaching curves I tried to do it by changing to polar coordinates: $x_1=r \cdot \sin \theta \cdot \cos \varphi $
$x_2=r \cdot \sin \theta \cdot \sin \varphi $
$x_3=r \cdot \cos \theta$

So you instead get the limit: $$\lim_{r\to 0^+}\frac {\ln(1+r^2)}{r^2+\sin (r^3\cdot \lambda)}$$ where $\lambda=\sin (\theta)^2\cos\theta\cos\varphi\sin\varphi$
By then using l'Hopitals rule I got:\begin{align}\lim_{r\to 0^+}\frac {\frac {2r}{1+r^2}}{2r + \lambda 3r^2\cos (r^3\cdot\lambda)}&=\lim_{r\to 0^+}\frac {1}{(1+r^2)\cdot (1 + 3\lambda r \cdot \cos(r^3\lambda))}\\&=\lim_{r\to 0^+}\frac {1}{1+r^2 + 3\lambda r \cdot \cos(r^3\lambda))+3\lambda r^3 \cdot \cos(r^3\lambda))}\\&=1\end{align}

If my approach is correct I wonder if I can solve this without l'Hopital maybe with equalities? .

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  • $\begingroup$ I think you mean $\lvert \bar{x} \rvert = \sqrt{x_{1}^{2} + x_{2}^{2} + x_{3}^{2}}$. Also, you can use Taylor expansions instead of L'Hopital. $\endgroup$ – Mattos Sep 29 '17 at 11:20
  • $\begingroup$ Yes indeed I did. $\endgroup$ – DrShellyCooper Sep 29 '17 at 11:24
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Just use the fact that $\lim\limits_{t\to0}\frac{\ln(1+t)}t = 1$ and $\lim\limits_{t\to 0} \frac{\sin t}{t} = 1$:

$$\lim_{r\to 0^+}\frac {\ln(1+r^2)}{r^2+\sin (\lambda r^3)} = \lim_{r\to 0^+}\frac {\ln(1+r^2)}{r^2}\frac{r^2}{r^2+\sin (\lambda r^3)} = \lim_{r\to 0^+}\frac {\ln(1+r^2)}{r^2}\frac{1}{1+\frac{\sin (\lambda r^3)}{r^2}} = 1 \cdot 1 = 1$$

because

$$\lim_{r\to 0^+}\frac{\sin (\lambda r^3)}{r^2} = \lim_{r\to 0^+}\frac{\sin (\lambda r^3)}{\lambda r^3}\cdot\lambda r = 1\cdot0 = 0$$

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