How to prove by definition (epsilon-delta) this limit of 2 variables?

Question

I have this limit $$\lim_{(x,y)\rightarrow(1,0)}x^2\cos(y)=1$$ How to prove by definition: $$\lim_{(x,y)\rightarrow(1,0)}x^2\cos(y)=L\Longleftrightarrow\forall\varepsilon>0,\exists\delta>0\ \big[ 0<\sqrt{(x-1)^2+(y-0)^2}<\delta\Rightarrow |x^2\cos(y)-1|<\varepsilon \big]$$ The trigonometric function is difficulting the usual computation of factoring.

Answer 1

Hint: Try writing

$$|x^2\cos y - 1| \leq |x^2\cos y - \cos y| + |\cos y - 1|$$ $$ \ \ \ \ \ \ \ \ \leq |x^2 - 1| + |\cos y - 1|$$

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Added:

The logic is the other way around: as $\cos y$ and $x^2$ are continuous, then for any $\epsilon > 0$ there exists $\delta_1, \delta_2 > 0$ such that

$$0 < |x - 1| < \delta_1 \ \Longrightarrow \ | x^2 - 1 | < \epsilon/2$$ and

$$ 0 < |y| < \delta_2 \ \Longrightarrow \ | \cos y - 1 | < \epsilon/2$$

Hence given $\epsilon > 0$, choose $\delta = \min(\delta_1,\delta_2)$, then

$$ \| (x,y) - (1,0) \| = \sqrt{ (x-1)^2 + y^2 } < \delta \ \Rightarrow 0 < |x - 1| < \delta \ \wedge 0 < |y| < \delta \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ $$

$$ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \Rightarrow 0 < |x - 1| < \delta_1 \ \wedge 0 < |y| < \delta_2 $$

$$ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \Rightarrow |x^2\cos y - 1| \leq |x^2 - 1| + |\cos y - 1| $$

$$ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ < \epsilon/2 + \epsilon/2 = \epsilon $$

Answer 2

Write $|x^2\cos(y)-1|$ as:

$$|x^2\cos(y)-1|=|(x^2-1)(\cos(y)-1)+x^2-1+\cos(y)-1| \leq \\ \leq |(x^2-1)||(\cos(y)-1)|+|x^2-1|+|\cos(y)-1|$$

Now you can write single variable delta-epsilon proof for $x^2-1$ and $\cos(y)-1$.