is my attempt correct? any help is appreciated.
Prove that the limit does not exist:
$$\lim_{x\to 0}(x+sgn(x))$$
I can prove that by letting
two sequences $x_n$: $1/n \to 0$ and $y_n$: $-1/n \to 0$, $x_n \neq 0$, $y_n \neq 0$, for all $n \in N$
$f(x_n) = (1/n+sgn(1/n)) = (1/n +1) \to 1$
$f(y_n)= (-1/n+sgn(-1/n)) = (-1/n +(-1)) \to -1$
since $f(x_n) \neq f(y_n)$, the limit D.N.E
thank you
There is another proof, albeit less rigorous. $$\lim\limits_{x\to0}(x+\operatorname{sgn}(x))=\lim\limits_{x\to0}(x)+\lim\limits_{x\to0}(\operatorname{sgn}(x))=0+\lim\limits_{x\to0}(\operatorname{sgn}(x))=\lim\limits_{x\to0}(\operatorname{sgn}(x))$$ In order for the ordinary limit to exist, both the left-side and the right-side limits as $x\to0$ must be equal. Now, $x\to0^-$ means $x\lt0$ and $x\to0^+$ means $x\gt0$. These are two of the three subdomains of the signum function. $$ \operatorname{sgn}(x)\overset{\text{def}}{=} \begin{cases} -1,&\text{if $x\lt0$}\\ 0,&\text{if $x=0$}\\ 1,&\text{if $x\gt0$} \end{cases} $$ Clearly, the left-side and right-side limits will not be equal. $$\lim\limits_{x\to0^-}(\operatorname{sgn}(x))=-1\ne1=\lim\limits_{x\to0^+}(\operatorname{sgn}(x))$$ As such, the original limit in this problem does not exist.