Here is Prob. 15, Sec. 5.1, in the book Introduction To Real Analysis by Robert G. Bartle and Donald R. Sherbert, 4th edition:

Let $f \colon (0, 1) \to \mathbb{R}$ be bounded but such that $\lim_{x \to 0} f$ does not exist. Show that there are two sequences $\left(x_n\right)$ and $\left(y_n\right)$ in $(0, 1)$ such that $\lim \left( x_n \right) = 0 = \lim \left( y_n \right)$, but such that $\lim \left( f \left(x_n\right) \right)$ and $\lim \left( f \left(y_n\right) \right)$ exist but are not equal.

My Attempt:

For each $n \in \mathbb{N}$, let us put $$ I_n \colon= \left( 0, \frac{1}{n } \right) = \left\{ \ x \in \mathbb{R} \ \colon \ 0 < x < \frac{1}{n} \ \right\}, \tag{0} $$ and hence let us take $$ \alpha_n \colon= \inf f \left( I_n \right) \qquad \mbox{ and } \qquad \beta_n \colon= \sup f \left( I_n \right). \tag{1} $$ Then as $I_{n+1} \subset I_n$, so we must have $$ \alpha_{n} \leq \alpha_{n+1} \leq \beta_{n+1} \leq \beta_n. \tag{2} $$ And, for any natural numbers $m$ and $n$ such that $m < n$, we see from (2) that $$ \alpha_m \leq \alpha_{m+1} \leq \cdots \leq \alpha_n \leq \beta_n, $$ and also $$ \alpha_n \leq \beta_n \leq \beta_{n-1} \leq \cdots \leq \beta_m. $$ Hence we can conclude without any loss of generality that, for any natural numbers $m$ and $n$, we have $$ \alpha_m \leq \beta_n. \tag{3} $$

Moreover, as our function $f$ is bounded on the open interval $(0, 1)$, so we must also have $$ -\infty < \inf f \big( (0, 1) \big) \leq \sup f \big( (0, 1) \big) < +\infty, \tag{4} $$ and in fact $$ \inf f \big( (0, 1) \big) \leq \alpha_n \leq \beta_n \leq \sup f \big( (0, 1) \big) \tag{5} $$ for all $n \in \mathbb{N}$.

Thus $\left( \alpha_n \right)$ is a monotonically increasing sequence of real numbers which is also bounded from above in $\mathbb{R}$, and $\left( \beta_n \right)$ is a monotonically decreasing sequence of real numbers which is also bounded from below in $\mathbb{R}$. Therefore both of these sequences are convergent in $\mathbb{R}$. Let us put $$ \alpha \colon= \lim_{n \to \infty } \alpha_n \qquad \mbox{ and } \qquad \beta \colon= \lim_{n \to \infty } \beta_n. \tag{6} $$ Then $$ \alpha \leq \beta, \tag{7} $$ by virtue of (3) above.

For each $n \in \mathbb{N}$, as $$ \alpha_n + \frac{1}{n} > \alpha_n \qquad \mbox{ and } \qquad \beta_n - \frac{1}{n} < \beta_n, $$ so (by the definition of the supremum and the infimum of a non-empty bounded subset of $\mathbb{R}$) there exist real numbers $x_n$ and $y_n$ in $I_n$ such that $$ \alpha_n \leq f \left( x_n \right) < \alpha_n + \frac{1}{n} \qquad \mbox{ and } \qquad \beta_n - \frac{1}{n} < f \left( y_n \right) \leq \beta_n. \tag{8} $$ [Refer to (0) and (1) above.]

Thus we have sequences $\left( x_n \right)$ and $\left( y_n \right)$ in the open interval $(0, 1)$ such that $$ 0 < x_n < \frac{1}{n} \qquad \mbox{ and } \qquad 0 < y_n < \frac{1}{n} $$ for all $n \in \mathbb{N}$. Therefore by the sandwiching theorem we can conclude that both of these sequences converge to $0$. But from (6) and (8) together with the sandwiching theorem we can also conclude that
$$ \lim_{n \to \infty} f \left( x_n \right) = \alpha \qquad \mbox{ and } \qquad \lim_{n \to \infty} f \left( y_n \right) = \beta. $$

Is what I've done so far correct? If so, then how to show that our $\alpha$ and $\beta$ in (6) above are different?

Or, have I made a mistake anywhere in my reasoning?

Or, is there any other (and easier and more direct) way of proving this result?


What you did is correct.

In order to show that $\alpha\neq\beta$, suppose otherwise. That is, suppose that $\alpha=\beta$. I will prove that $\lim_{x\to0}f(x)=\alpha(=\beta)$, thereby reaching a contradiction. Take $\varepsilon>0$. Now, take $n\in\mathbb N$ such that $\beta_n-\alpha_n<\varepsilon$; it must exist, since $\lim_{n\to\infty}\beta_n-\alpha_n=\beta-\alpha=0$. But then, by the definition of $\alpha_n$ and $\beta_n$,$$x\in\left(0,\frac1n\right)\implies\alpha_n\leqslant f(x)\leqslant\beta_n\implies\bigl|f(x)-\alpha\bigr|<\varepsilon,$$since both $f(x)$ and $\alpha$ belong to $(\alpha_n,\beta_n)$ and $\beta_n-\alpha_n<\varepsilon$.


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