# can we say a function $f:[a,b]\to \Bbb{R}$ is continuous at the end points?

$$f:[a,b]\to \Bbb{R}$$ 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?

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.

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 $$a,b$$.

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