For my vector calculus course, I need to solve this limit: $$\lim_{(x,y)\rightarrow (0,0)} \frac{1-\cos(x^2y)}{x^6+y^4}$$
By replacing (x,y) by (0,0), I noticed that the limit was of the form $\frac{0}{0}$ and decided to Taylor expand the numerator which gave me this inequality: $$\left | 1-\cos(x^2y) \right |\leq\frac{x^4y^2}{2}$$
Then I rewrote my limit using the inequality I just found: $$\left | \frac{1-\cos(x^2y)}{x^6+y^4} \right |\leq \frac{x^4y^2}{2(x^6+y^4)}\leq \frac{x^4y^2}{x^6+y^4}$$
Seeing that I was stuck, I decided to rewrite my expression in polar coordinates: $$\left | \frac{1-\cos(x^2y)}{x^6+y^4} \right |\leq\frac{(r\cos(\theta))^4(r\sin(\theta))^2}{(r\cos(\theta))^6+(r\sin(\theta))^4}$$
By making some simplifications, I get: $$\frac{(r\cos(\theta))^4(r\sin(\theta))^2}{(r\cos(\theta))^6+(r\sin(\theta))^4}=\frac{r^2\cos^4(\theta)\sin^2(\theta)}{r^2\cos^6(\theta)+\sin^4(\theta)}$$
And when I take the limit of my new expression when r tends to 0, I get: $$\lim_{r\to 0} \frac{r^2\cos^4(\theta)\sin^2(\theta)}{r^2\cos^6(\theta)+\sin^4(\theta)} = \frac{0}{\sin^4(\theta)}=0$$ unless $\theta=k\pi$ where $k \epsilon \mathbb{Z}$
First, I wanted to know if my proof was correct and if it is not, where was my mistake. Second, I wanted to know if there was a proof without the condition that $\theta=k\pi$.
Thank you in advance for your answers,
IsaacM
One possible variant to do is, that knowing $1-\cos \alpha=2\sin^2 \frac{\alpha}{2}$ we have $$\frac{1-\cos(x^2y)}{x^6+y^4} = \frac{2\sin^2 \frac{x^2y}{2}}{x^6+y^4}$$ So we can consider $$\frac{x^4y^2}{x^6+y^4} \leqslant \frac{x^4y^2}{2|x|^3y^2}=\frac{1}{2}x\cdot\text{sign}(x)\to 0$$