3
$\begingroup$

$f:[a,b]\to \Bbb{R}$ is said be continuous at a point $m$ that means whenever a sequence $x_n \in[a,b]$ converge to ${m}$ the image ${f(x_n)} $ converge to ${f(m)}$ and by applying this definition of continuous function if $x_n \in[a,b]$ converge to ${a}$ the image ${f(x_n)}$ converge to ${f(a)}$ and that means that the function is continuous at $a$

but if we apply the limit definition of continuity clearly $\lim_{x\to a^+}f(x)$ does not equal $\lim_{x\to a^-}f(x)$ because the limit doesn't exist in the left side so is the function continuous at its end points? and if so what did I get wrong?

$\endgroup$

3 Answers 3

7
$\begingroup$

Strictly speaking the limit definition of continuity is not $\lim_{x \to a^+} f(x) = \lim_{x \to a^-} f(x) = f(a)$. That itself is a special case derived when both the quantities $\lim_{x \to a^+} f(x)$ and $\lim_{x \to a^-} f(x)$ make sense.

The limit definition of continuity for $f$ at $a$ is that $\lim_{x \to a} f(x) = f(a)$. If you want to be explicit: for all $\varepsilon > 0$, there is $\delta > 0$ s.t. $$ x \in (a - \delta, a + \delta) \cap [a, b] \implies |f(x) - f(a)| < \varepsilon $$ In this case since $a$ is the left endpoint of $[a, b]$, $(a - \delta, a + \delta) \cap [a, b] = [a, a + \delta) \cap [a, b]$ and the above condition collapses to requiring $$ x \in [a, a + \delta) \cap [a, b] \implies |f(x) - f(a)| < \varepsilon $$ i.e. the requirement $\lim_{x \to a} f(x) = f(a)$ is identical to $\lim_{x \to a^+} f(x) = f(a)$ in this case.

$\endgroup$
1
$\begingroup$

Continuity definition is w.r.t subspace topology i.e., w.r.t based on limits of $(x_k:k)$ inside $[a,b]$. So yes you can say the function $f$ is continuous at $a,b$ as long as $f$ is defined in $[a,b]$. But now if you extend the function $f$ to a larger domain then you need to check again if $f$ is continuous at $a,b$.

$\endgroup$
1
$\begingroup$

For function to be continuous on an interval $[a,b]$ it has to be:

  1. continuous on whole interval $(a,b)$

  2. $\lim_{x\to a^+}{f(x)}=f(a)$, because $f$ isn't defined for $x \notin [a,b]$, so we can't look at left limit $x$ if it isn't defined on that interval

  3. $\lim_{x\to b^-}{f(x)}=f(b)$, similarly to the $2.$ condition

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .