This problem is an example in my calculus textbook. Let: $$ f(x,y)=\frac{y^2\sin^2(x)}{x^4+y^4} $$ My textbook says that $$ \lim_{(x,y)\to(0,0)}f(x,y) \text{ does not exist.} $$ Questions:
How do we know the limit does not exist?
In general, suppose the limit of a function exists but we do not know the value of such limit, how do we find it?
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1: Use the approximation (for small $x), \ sin(x)\approx x$. Then $f(x,y)\approx \frac{x^2y^2}{x^4+y^4}$. Now consider two cases:
(a) $x=y$ then the function $=\frac{1}{2}$.
(b) $x=0$ or $y=0$ where the function $=0$, when the other variable $\ne 0$.
Therefore the limit does not exist.
2: There is no general solution. Each problem must be evaluated on its own.
Consider the curves $\gamma_{1} = (t,t)$ and $\gamma_{2}(t) = (t,t^{2})$. Thus we have \begin{align*}f(\gamma_{1}(t)) = \frac{t^{2}\sin^{2}(t)}{t^{4} + t^{4}} = \frac{\sin^{2}(t)}{2t^{2}} \Longrightarrow \lim_{t\rightarrow 0}f(\gamma_{1}(t)) = \lim_{t\rightarrow 0}\left[\frac{1}{2}\left(\frac{\sin(t)}{t}\right)^{2}\right] = \frac{1}{2} \end{align*}
On the other hand, we have \begin{align*} f(\gamma_{2}(t)) = \frac{t^{8}\sin^{2}(t)}{t^{4} + t^{8}} = \frac{t^{4}\sin^{2}(t)}{1 + t^{4}} \Longrightarrow \lim_{t\rightarrow 0}f(\gamma_{2}(t)) = \lim_{t\rightarrow 0}\frac{t^{4}\sin^{2}(t)}{1+t^{4}} = \frac{0\times 0}{1+0} = 0 \end{align*}
If the given limit existed, we should have $\displaystyle\lim_{t\rightarrow0}f(\gamma_{1}(t)) = \lim_{t\rightarrow 0}f(\gamma_{2}(t))$. Hence $\displaystyle\lim_{\textbf{x}\rightarrow\textbf{0}} f(\textbf{x})$ does not exist. Hope this helps.