# proving a function is differentiable from the right

This question is asking about understanding a solution to a problem. Below is the solution. The thing I'm not understanding is why showing that if $$\lim\limits_{n\to\infty} g(x_n)$$ exists for every sequence $$\{x_n\}$$ in $$(0,1)$$ converging to $$0$$, then $$\lim\limits_{x\to 0^+} g(x)$$ exists. I know that if I can show that $$\lim\limits_{n\to\infty} g(x_n)$$ exists for every sequence $$\{x_n\}$$ in $$(0,1)$$ with $$x_n\to 0$$ and that this limit is unique, then if $$\lim\limits_{x\to 0^+} g(x)$$ does not exist, it cannot be equal to this unique value, say $$L$$. Then by definition,we may find $$\epsilon > 0$$ so that $$\forall n\in\mathbb{Z}^+,\exists x_n \in (0, \frac{1}n)$$ so that $$|g(x_n) - L| \ge \epsilon$$, and this contradicts the fact that we have $$g(x_n)\to L$$. But how can I show that for every sequence $$(x_n)\subseteq (0,1), x_n\to 0, g(x_n)\to L$$? Is this even true?

If $$g(x_n)\rightarrow L_1$$ and $$g(y_n)\rightarrow L_2$$, and define $$z_{2n}=x_n$$ and $$z_{2n+1}=y_n$$, then $$\limsup g(z_n)=\max(L_1,L_2)$$ and $$\liminf g(z_n)=\min(L_1,L_2)$$. But, by hypothesis, these two quantities are equal, otherwise $$g(z_n)$$ would not converge; and therefore $$L_1=L_2$$. So all sequences converge to the same limit as well.