Squeeze theorem
Fourth example

The squeeze theorem can still be used in multivariable calculus but the lower (and upper functions) must be below (and above) the target function not just along a path but around the entire neighborhood of the point of interest and it only works if the function really does have a limit there. It can, therefore, be used to prove that a function has a limit at a point, but it can never be used to prove that a function does not have a limit at a point.

lim (x, y) → (0, 0) x 2 y x 2 + y 2 {\displaystyle \lim _{(x,y)\to (0,0)}{\frac {x^{2}y}{x^{2}+y^{2}}}}

cannot be found by taking any number of limits along paths that pass through the point, but since

0 ≤ x 2 x 2 + y 2 ≤ 1 − | y | ≤ y ≤ | y | ⟹ − | y | ≤ x 2 y x 2 + y 2 ≤ | y | lim (x, y) → (0, 0) − | y | = 0 lim (x, y) → (0, 0) | y | = 0 ⟹ 0 ≤ lim (x, y) → (0, 0) x 2 y x 2 + y 2 ≤ 0 {\displaystyle {\begin{array}{rccccc}&0&\leq &\displaystyle {\frac {x^{2}}{x^{2}+y^{2}}}&\leq &1\\-|y|\leq y\leq |y|\implies &-|y|&\leq &\displaystyle {\frac {x^{2}y}{x^{2}+y^{2}}}&\leq &|y|\\{{\displaystyle \lim _{(x,y)\to (0,0)}-|y|=0} \atop {\displaystyle \lim _{(x,y)\to (0,0)}\ \ \ |y|=0}}\implies &0&\leq &\displaystyle \lim _{(x,y)\to (0,0)}{\frac {x^{2}y}{x^{2}+y^{2}}}&\leq &0\end{array}}}

therefore, by the squeeze theorem,

lim (x, y) → (0, 0) x 2 y x 2 + y 2 = 0. {\displaystyle \lim _{(x,y)\to (0,0)}{\frac {x^{2}y}{x^{2}+y^{2}}}=0.}
keywords:existence