Relation between delta-epsilon when proving limits

Question

Assume we know the following about the function $h$:

$|x − 1| < 0.2 =⇒ |h(x) − 4| < 1.$

Can you guarantee that there exists $δ > 0$ such that (is it possible) $|x − 1| < δ =⇒ |h(x) − 4| < 0.5?$

And what about this $|x − 1| < δ =⇒ |h(x) − 4| < 1.5$?

Answer 1

If $h(x)$ is the constant function $h(x)=4.9$, the given condition is satisfied, but no choice of $\delta$ will ever put $|h(x)-4|=\frac9{10}<\frac12$.

On the other hand, $|h(x)-4|<1\implies |h(x)-4|<1.5$, because $1<1.5$, so that part's not a problem.

Answer 2

Limit of a function:

For every $\epsilon \gt 0,$ there exists a $\delta \gt 0,$

such that $|x-a| \lt \delta$ implies $|h(x)-L| \lt \epsilon$.

$\iff : $

For every $(x_n)_{n \in \mathbb{N}},$

with $\lim _{n \rightarrow \infty} x_n = a,$ $y_n:= h(x_n) $

we have:

$\lim _{n \rightarrow \infty } y_n$ exists and equals $L.$

1) If $h(x)$ is continuos at $x= a,$ then

the limit exists and equals $L = h(a).$

2) The problem arises when lim $y_n$ does not exist, or exists and $\ne L.$

In which category do you put Tony's example?

Can you find an example where $\lim y_n$

does not exist and :

There exists a $n_0$ such that for $n \ge n_0 :$

$|y_n - 4| \lt 1$ ?