Is this family of functions equicontinuous in $\mathbb{R}$? Let $\{ \varphi_{\lambda}(x,y): \lambda \in \mathbb{R} \}$ be the family of functions defined by $$\varphi_{\lambda}(x,y):= \frac{1}{1 + x^4 + \frac{1}{2}\arctan(\lambda)\sin(y^6)}$$ for all $\lambda \in \mathbb{R}$. Is this family of functions equicontinuous in $\mathbb{R}$?

It is easy to see that this family of functions is uniformly bounded by $1$, but I don't know how to see if is equicontinuous. I have thought about using subsequences and maybe Arzela-Ascoli theorem, but I do not get the point... For example, it is not equi-Lipschitz as its partial derivatives are not bounded in a convex open set, so this doesn't prove nothing neither...
Any hint will be appreciated. Thanks in advance
 A: Take $\lambda,\mu$.
Since $\frac1a-\frac1b = \frac{b-a}{ab}$ we have
$$
|\phi_\lambda(x,y)-\phi_\mu(x,y)|\le \frac1{(1-\frac\pi4)^2} \frac12 |\arctan(\lambda)-\arctan(\mu)|.
$$
This implies $\phi_\mu\to \phi_\lambda$ uniformly for $\arctan(\mu) \to \arctan(\lambda)$.
Since the set $\{\arctan \lambda: \ \lambda\in \mathbb R\}$ is bounded, we can find for each  sequence $(\lambda_n)$ a subsequence such that $(\arctan(\lambda_{n_k}))$  converges, and so $(\phi_{\lambda_{n_k}})$ converges uniformly. Then by Arzela-Ascoli, the family is equicontinuous.
A: A direct calculation/estimate works: The denominator is bounded below by $1 - \pi/4 > 0$, so that
$$
| \varphi_{\lambda}(x_1,y_1) - \varphi_{\lambda}(x_2,y_2)|
\le  \frac{|x_2^4 + \frac{1}{2}\arctan(\lambda)\sin(y_2^6) - x_1^4 - \frac{1}{2}\arctan(\lambda)\sin(y_1^6)|}{(1-\pi/4)^2} \\
\le \frac{|x_2^4 -x_1^4|+ \frac{\pi}{4}|\sin(y_2^6)-\sin(y_1^6)|}{(1-\pi/4)^2} = C |f(x_2)-f(x_1)| + D|g(y_2)-g(y_1)|
$$
with some constants $C, D$ and the continuous functions $f(x) = x^4$ and $g(y) = \sin(y^6)$.
For given $(x_1, y_1) \in \Bbb R^2$ this becomes arbitrary small if $(x_2, y_2)$ is sufficiently close to $(x_1, y_1)$, independently of $\lambda$.
