I am trying to find the limit of the function
$$ f(x,y)=\frac{1}{x}\sin\frac{xy^2}{\sqrt{x^2+y^2}} $$
for $(x,y)$ approaching $(0,0)$. Based on the $\frac{1}{x}$ I am expecting for the limit to not exist. So I have tried to approach along the different axes.
For the approach along the $x$-axis, I get a limit value of $0$.
However, when I approach along the $y$-axis, I end up with the following limit:
$$ \lim_{y\to0} (\frac{1}{0}\sin(0y))$$ I am inclined to say that this limit does not exist, as I cannot divide by zero. Is this a correct assumption? And can I then conclude that the given limit does not exist?
Best Answer
To begin with, note that $$ \lim_{(x,y)\to (0,0)}\frac{x\,y^2}{\sqrt{x^2+y^2}}=\lim_{(x,y)\to (0,0)}\sqrt\frac{x^2}{x^2+y^2}\,y^2=0 $$ because the square root is bounded. Then you can use the Taylor expansion of the $\sin$: $$ \lim_{(x,y)\to (0,0)}\frac{1}{x}\sin\frac{x\,y^2}{\sqrt{x^2+y^2}}=\lim_{(x,y)\to (0,0)}\frac{1}{x}\frac{x\,y^2}{\sqrt{x^2+y^2}}=\lim_{(x,y)\to (0,0)}\sqrt\frac{y^2}{x^2+y^2}\,y=0 $$ since again the root is bounded.
To explain that apparently the limit does not exist because you cannot approach the origin by $x=0$, note that the limit when $x\to 0,y\ne 0$ does exist. So the extension by continuity of $f$ (considering only the variable $x$ and a constant $y\ne 0$) is well defined at any point but $(0,0)$.