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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?

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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.

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