I have to prove that $\displaystyle\lim_{(x,y)\to(1,0)} \frac{(x-1)^3\cos{y}}{(x-1)^2+y^2}=0$ using the epislon-delta definition of a limit.
Here's what I figured:
$\delta>|x-1| \Rightarrow \delta>(x-1)^2\Rightarrow \delta^3>|(x-1)^3|$. Also, $\delta>|y|\geq|cos{y}| \Rightarrow \delta^2>y^2$. So $\delta^4>|(x-1)^3\cos{y}|$ and $2\delta^2>|(x-1)^2+y^2|$. Obviously, $\delta^4>|\frac{(x-1)^3\cos{y}}{(x-1)^2+y^2}|$ if $(x-1)^2+y^2\geq1.$ But how do I account for it otherwise? I can't think of anything. I'm having issues proving the limits of quotients in general.
Thank you for your help!
If you denote $$f(x,y)=\frac{(x-1)^3\cos{y}}{(x-1)^2+y^2},$$ notice that
$$\vert f(x,y) \vert = \left\vert \frac{(x-1)^2}{(x-1)^2+y^2} \right\vert \vert \cos y \vert \vert x-1 \vert \le \vert x-1 \vert.$$
Then taking any $\epsilon \gt 0$, you have for $\delta = \epsilon$ and $\vert x - 1 \vert \lt \delta$:
$$\vert f(x,y) \vert \le \vert x - 1 \vert \lt \delta = \epsilon$$
which concludes the proof.