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$f:\mathbb{R}^2 \setminus \{(x,y): xy =0 \}, f(x,y)= (x+y)\sin{\frac1{x}}\cos{\frac1{y}}$

I have to evaluate limit $$\lim_{(x,y) \to (0,0)} f(x,y)$$

By sandwich theorem I get that limit is $0$.$$0 \le |f(x,y)| = (|x+y|) \left | \sin{\frac1{x}} \cos{\frac1{y}} \right | \le|x| + |y|$$

So my question is how can $\lim_{(x,y) \to (0,0)} f(x,y)$ exist if limits $\lim_{(x \to 0)} f(x,y)$ and $\lim_{(y \to 0)} f(x,y)$ don't exist.

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    $sin(\frac 1y)$ or $cos(\frac 1y)$? – 2017-02-09

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If you fix $y \neq 0$ then indeed $\displaystyle \lim_{x \to 0} f(x,y)$ does not exist. What makes you think that $$ \lim_{(x,y) \to (0,0)} f(x,y) \text{ exists} \implies \lim_{x \to 0} f(x,y) \text{ exists } \forall y $$ For fixed $y \neq 0$, it is not the case that if $x \to 0$ then $(x,y) \to (0,0)$. In fact one has $(x,y) \to (0,y)$ in that case.