Proving multivariable limit using epsilon-delta definition Let $\displaystyle g(x, y) = \frac{\sin^2(x - y)}{|x| + |y|}$. Prove
  that $\lim_{(x, y)\to(0,0)} g(x) = 0$.


Any help on this would be appreciated. So far I think that the fact that $|x + y| \le |x| + |y|$ and $|\sin(x + y)| \le |x + y|$ has to be utilized but I am not sure how to do so in the epsilon delta proof.
Remember that $|\sin(z)| &lt; z$ for $z \ne 0$.  So $0 \le \sin^2(z) &lt; z^2$.  Set $z = x-y$.  Your denominator satisfies $|z| = |x-y| \le |x| + |y|$ by the triangle inequality.

$$0 \le \lim_{(x,y) \to (0,0)} \frac{\sin^2(x-y)}{|x|+|y|} \le \lim_{(x,y) \to (0,0)} \frac{z^2}{|z|} = \lim_{(x,y) \to (0,0)} |z| = 0$$

You can easily convert this proof, which uses the squeeze theorem, into an $\epsilon-\delta$ proof by noting the inequalities and setting $\delta = \epsilon$.  For $\Vert ( x,y ) \Vert &lt; \epsilon$, we have $|z| &lt; \epsilon$ since $|x-y| \le \sqrt{x^2 + y^2}$.
An idea:

$$|x-y|\le|x|+|y|\implies \frac{\sin^2(x-y)}{|x|+|y|}\le\frac{\sin^2(x-y)}{|x-y|}=|\sin(x-y)|\frac{|\sin(x-y)|}{|x-y|}$$

Now just remember that $\,\displaystyle{\lim_{t\to 0}\frac{\sin t}t=1}\;$ and stuff...