Prove that $\frac{\sin(xy + y^3)}{x^2 + y^2}$ has no limit at the origin

Question

I have this function, $$ \frac{\sin(xy + y^3)}{x^2 + y^2} $$ I have to prove that for $(x, y)\to(0, 0)$, limit doesn't exist, and because of that, function is not continuous (i.e. function diverges).

I've used polar coordinates method, in the denominator I've got only $\rho^2$ because of trig identities (i.e. $\sin^2\theta+ \cos^2\theta= 1$) overall, I've obtained this function, $$(1/\rho^2)\sin(\dots)$$

I've used this inequality, $$|1/\rho^2||\sin(...)|\leq|1/\rho^2|$$ in the right hand side I didn't write $\sin(\dots)=$ because I took the maximum value of sin, (i.e. $1$) because it's a limited function ranging from $-1$ to $1$ (the absolute value of sin is ranging from $0$ to $1$). now in the right hand side there's a $g(\rho)$ function, and $g(\rho)$ depends only on $\rho$. Now, for $\rho\to0$, $g(\rho^2)$ tends to $+\infty$.

Based on this conclusion, can I say the function in the left hand side tends to $+\infty$ too (i.e. diverges)? or, is polar coordinares method enough to prove divergence of this kind of limits?

Answer 1

OK, this answer has two parts. First, I will explain why what you did was wrong, then I will give you just a hint how to proceed correctly. I suggest you read both parts.

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Part 1:

What you are suggesting is the following logic:

• Prove that, for every $\rho$ the function $f(\rho,\phi)$ is smaller than some other function $g(\rho)$.
• Prove that $\lim_{\rho\to 0} g(\rho)=\infty$.
• From there, conclude that $\lim_{\rho\to 0} f(\rho,\phi)=\infty$.

The reasoning above is incorrect. The conclusion does not follow from the two steps preceeding it. For example, if the reasoning above were correct, then I could prove that the function $f(x,y)=0$ is also not continuous at $0$:

• Obviously, $f(\rho,\phi) \leq \frac1\rho$.
• Clearly, $\lim_{\rho\to 0} \frac1\rho = \infty$.
• Therefore, $\lim_{\rho\to 0}f(\rho,\phi)=\infty$.

The conclusion is absurd, because the reasoning is faulty.

Remember: If a function $f$ is bigger than some function $g$, then the limit of $f$ must be bigger than the limit of $g$. But if the function is **smaller, then the limit must also be smaller, and if the limit of $g$ is $\infty$, then the reasoning tells you nothing about $f$.

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Part 2:

To actually solve the problem, I suggest you look at the limit of the function along a couple of lines. For example, the limit along the line $y=0$ should be relatively easy to calculate.

Now all you need to do is to find some other line along which the limit will not be the same. I suggest you try to find another simple straight line along which the expression will greatly simplify.

Answer 2

Perhaps

$$\frac{\sin(xy+y^3)}{x^2+y^2}=\frac{\sin(xy+y^3)}{xy+y^3}\cdot\frac{xy+y^3}{x^2+y^2}$$

Observe that $\;xy+y^3\xrightarrow[(x,y)\to(0,0)]{}0\;$ and thus the first factor on the right side converges to $\;1\;$ , yet for the second factor we get the limit doesn't exist because

$$\frac{xy+y^3}{x^2+y^2}=\begin{cases}\frac{x^2+x^3}{2x^2}=\frac{1+x}2\xrightarrow[x\to0]{}\frac12\,,\,\,when\;\;y=x\,,\,x\to0\\{}\\\frac0{x^2}=0\xrightarrow[x\to0]{}0\;,\;\;when\;\;y=0\,,\,x \to0\end{cases}$$

and thus the original limit doesn't exist.

Answer 3

$$\lim_{(x,y)\to (0,0)} \frac{\sin(xy + y^3)}{x^2 + y^2} $$ Consider the path $y=x$, and recall that $\sin t\sim t $ as $t\to0$: $$f(x,x)=\frac{\sin(x^2 + x^3)}{2x^2 }\to \frac{1}{2}\quad \text{as}\quad x\to 0 $$ Therefore the limit does not exist.