Vtyjkyuk

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 $f colon (0, 1) to mathbbR$ be bounded but such that $lim_x to 0 f$ does not exist. Show that there are two sequences $left(x_nright)$ and $left(y_nright)$ in $(0, 1)$ such that $lim left( x_n right) = 0 = lim left( y_n right)$, but such that $lim left( f left(x_nright) right)$ and $lim left( f left(y_nright) right)$ exist but are not equal.

My Attempt:

For each $n in mathbbN$, let us put
$$ I_n colon= left( 0, frac1n right) = left x in mathbbR colon 0 < x < frac1n right, tag0 $$
and hence let us take
$$ alpha_n colon= inf f left( I_n right) qquad mbox and qquad beta_n colon= sup f left( I_n right). tag1 $$
Then as $I_n+1 subset I_n$, so we must have
$$ alpha_n leq alpha_n+1 leq beta_n+1 leq beta_n. tag2 $$
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
$$ alpha_m leq beta_n. tag3 $$

Moreover, as our function $f$ is bounded on the open interval $(0, 1)$, so we must also have $$ -infty < inf f big( (0, 1) big) leq sup f big( (0, 1) big) < +infty, tag4 $$
and in fact
$$ inf f big( (0, 1) big) leq alpha_n leq beta_n leq sup f big( (0, 1) big) tag5 $$
for all $n in mathbbN$.

Thus $left( alpha_n right)$ is a monotonically increasing sequence of real numbers which is also bounded from above in $mathbbR$, and $left( beta_n right)$ is a monotonically decreasing sequence of real numbers which is also bounded from below in $mathbbR$. Therefore both of these sequences are convergent in $mathbbR$. Let us put
$$ alpha colon= lim_n to infty alpha_n qquad mbox and qquad beta colon= lim_n to infty beta_n. tag6 $$
Then $$ alpha leq beta, tag7 $$
by virtue of (3) above.

For each $n in mathbbN$, as
$$ alpha_n + frac1n > alpha_n qquad mbox and qquad beta_n - frac1n < beta_n, $$
so (by the definition of the supremum and the infimum of a non-empty bounded subset of $mathbbR$) there exist real numbers $x_n$ and $y_n$ in $I_n$ such that
$$ alpha_n leq f left( x_n right) < alpha_n + frac1n qquad mbox and qquad beta_n - frac1n < f left( y_n right) leq beta_n. tag8 $$
[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
$$ 0 < x_n < frac1n qquad mbox and qquad 0 < y_n < frac1n $$
for all $n in mathbbN$. 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 $alphaneqbeta$, suppose otherwise. That is, suppose that $alpha=beta$. I will prove that $lim_xto0f(x)=alpha(=beta)$, thereby reaching a contradiction. Take $varepsilon>0$. Now, take $ninmathbb N$ such that $beta_n-alpha_n<varepsilon$; it must exist, since $lim_ntoinftybeta_n-alpha_n=beta-alpha=0$. But then, by the definition of $alpha_n$ and $beta_n$,$$xinleft(0,frac1nright)impliesalpha_nleqslant f(x)leqslantbeta_nimpliesbigl|f(x)-alphabigr|<varepsilon,$$since both $f(x)$ and $alpha$ belong to $(alpha_n,beta_n)$ and $beta_n-alpha_n<varepsilon$.