Exercise · ZW-MULTIVARIABLE-LIMIT-004 · ★★☆ Intermediate
Compute \(\displaystyle\lim_{(x,y)\to(0,0)}\dfrac{x^2y}{x^2+y^2}\).
Choose one option.
A \(\dfrac12\) B \(0\) C \(1\) D Does not exist
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Explanation Since \(x^2\le x^2+y^2\), we have \(\left|\dfrac{x^2y}{x^2+y^2}\right|\le|y|\to0\); by the squeeze theorem the limit is \(0\).
The limit of a function \(f\) at a point \(\mathbf a\) is the number \(L\) that the values \(f(\mathbf x)\) approach as \(\mathbf x\) gets close to \(\mathbf a\) along any path.
\[\lim_{\mathbf x\to\mathbf a} f(\mathbf x)=L\]
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