Find the points of discontinuities of the function $x \sin\left(\frac{\pi}{x^2+x}\right)$ Need to find the points of discontinuities of the function $f(x) = x \sin\left(\dfrac{\pi}{x^2+x}\right)$
Obviously at zero. I can’t take the one-sided limit ($\lim{x \to 0+}$ and $x \to 0-$) any way, sin is periodic, and at zero has a million oscillations.
In zero discontinuities, how to prove it. And how to get similar limits.


 sorry for lang. Frrrrom Rrrussia)

 A: Actually, you are able to take the limit at $0$. Note that $-1\leq\sin(y)\leq 1$ for any $y$, so we have:
$$-\vert x\vert \leq x\sin(\frac{\pi}{x^{2} + x})\leq \vert x \vert\text{ for }x\neq 0$$
We have the $x\neq 0$ condition because we wish to inspect the function's behavior at $x=0$. Then, we have:
$$\lim_{x\rightarrow 0}-\vert x\vert\leq\lim_{x\rightarrow 0}x\sin(\frac{\pi}{x^{2} + x})\leq \lim_{x\rightarrow 0}\vert x\vert$$
$$0\leq\lim_{x\rightarrow 0}x\sin(\frac{\pi}{x^{2} + x})\leq 0$$
$$\lim_{x\rightarrow 0}x\sin(\frac{\pi}{x^{2} + x}) = 0$$
To prove that the limit exists, you can do the same thing with left- and right-sided limits. You will get that they are both $0$. This method of evaluating limits is called the squeeze theorem and it is useful when you want to find the limit of a messy trig function.
However, this function is still not continuous at $x=0$ because it cannot be evaluated there.
A: It's not true that you can't take the limit as $x \to 0$.
Actually, using the usual bound $$-1 \le \sin (\ \text{something} \ ) \le 1$$
you get
$$-|x| \le f(x) \le |x|$$
Hence by the squeeze theorem
$$\lim_{x \to 0} f(x)=0$$
A: Actually, $f$ is continuous at $0$, since, for each $f$ in the sdomain of $f$, $|f(x)|\leqslant|x|$, and therefore $\lim_{x\to0}f(x)=0=f(x)$. Being continuous has nothing to do with oscillationg.
And, in fact, your function is continuous at every point of its domain (assuming that the domain is $\Bbb R\setminus\{-1\}$ and that $f(0)=0$).
A: The function is defined when $x^2+x\ne0$, so the domain is $\Bbb R-\{0,-1\}$. It is continuous over its domain. It would be wrong to say that the function is "discontinuous" at $0$ since $0$ does not belong to the domain of the function, so $f(0)$ is not defined.
We can however talk about the limit of the function at $0$.$$L=\lim_{h\to0}h\sin\left(\frac\pi{h^2+h}\right)$$ Note that $\sin$ is bounded within $[-1,1]$ irrespective of what is inside $\sin$. So as $h\to0$, the function approaches $0$, giving $L=0$. Alternatively note that $|f(x)|\le|x|$ and use the squeeze theorem to get $0\le\lim |f(x)|\le\lim|x|=0$.
A: Given $f(0)=0$, $f(x)$ will be continuous as the left and right limit abiut $x=0$ exist by sandwich theorem:$$ -x\le x\sin \frac{\pi}{x^2+x} \le x ~~~~(1)$$.
But by sandwich theorem (1) left lim is 1 and right limit is -1 about $x=-1$
the limit does not exist.So $f(x)$ is essentially discontinuous at $x=-1$.
