# real analysis: prove the limit does not exist

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

• Please use latex formatting! "\lim" instead of "lim", latex missing in the last lines ... Commented Apr 25, 2020 at 4:56
• Yes, what you have done is correct. Commented Apr 25, 2020 at 5:07
• Ok thank you for your time. Commented Apr 25, 2020 at 5:12

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.
• @sai-kartik We can separate the original ordinary limit into its two one-side limits. As such, it is permissible to separate $x$ and $\operatorname{sgn}(x)$ as the limits for those terms exist for both one-side limits. Additionally, the original function is continuous everywhere but $x=0$, but the value of $x$ only approaches $0$; it will not equal it. This is not a misuse of the limit law. Commented Jul 25, 2020 at 7:07