# If $\lim_{x \to x_0} f(x) = f(x_0)$, as $x \to x_0$, can we always say $f(x) \to f(x_0)$

Assuming $$\lim_{x \to x_0} f(x) = f(x_0)$$

As $$x$$ approaches $$x_0\$$i.e. $$(x\to x_0)$$, can we always say $$f(x)$$ approaches $$f(x_0)\$$ i.e. $$(f(x)\to f(x_0))$$? From what I understand, this is the motivation for the $$\epsilon$$-$$\delta$$ definition of limit.

The $$\epsilon-\delta$$ definition states that- $$\forall \epsilon$$, $$\exists \delta$$, such that:

$$0<|x-x_0|<\delta\Rightarrow|f(x)-L|<\epsilon$$. For any value of $$\epsilon$$ if I can provide a $$\delta$$, then $$L$$ is the limit of $$f(x)$$ at $$x_0$$.

Approach is apparently an informal idea, and not what the limit defines. So the idea of $$f(x)$$ getting closer to $$f(x_0)$$, as $$x$$ gets closer to $$x_0$$ is inaccurate and not always true.

Update: The statement apparently is true, when $$f(x)$$ is continuous at $$x_0$$. Though I have been unable to find a general proof for it yet.

• Why this wouldn't be true for constant functions? Commented Mar 15, 2021 at 2:54
• @CélioAugusto, this is just my understanding but, if $f(x)=a$, then for any value of $x$, $f(x)$ is always constant. If I'm trying to find its limit at a point $x_0$, as $x$ approaches $x_0$, I cannot say $f(x)$ approaches $a$. Commented Mar 15, 2021 at 11:12
• sure you can. $\epsilon - \delta$ definition doesn't fail in your example. The word approach may imply some kind of motion in some sense, but if you think about it there isn't any better approach of $a$ than itself. Commented Mar 15, 2021 at 11:50
• @alphaomega, if I don't think of approach as some kind of change occurring, I can probably say that. Thanks. Commented Mar 15, 2021 at 17:13
• Also as long as $\epsilon-\delta$ is satisfied, can we say for any function there is an approach? Commented Mar 15, 2021 at 17:14

Yes. If $$\lim_{h\to 0}f(x_0+h)=L$$ then for every $$\varepsilon>0$$ there exists $$\delta>0$$ such that whenever $$|h|<\delta$$ then $$|f(x_0+h)-L|<\varepsilon$$.
This means, for the same value of $$\varepsilon$$ and $$\delta$$, if $$|x-x_0|<\delta$$ then, setting $$h=x-x_0$$ we have $$|f(x)-L|=|f(x_0+h)-L|<\varepsilon$$, so the definition of $$f(x)\to L$$ as $$x\to x_0$$ is satisfied.
Note that this is actually slightly stronger than saying $$f(x)\to L$$ as $$x\to x_0$$, as $$\lim_{h\to 0}f(x_0+h)=L$$ also implies that $$f(x_0)=L$$, whereas in the definition of "$$f(x)\to L$$ as $$x\to x_0$$" it doesn't matter whether $$f(x_0)$$ even exists (or, if it does, what value it takes).
• How does $\lim_{h\to 0}f(x_0+h)=L$, mean $L=f(x_0)$? $f(x_0)$ could be undefined. Also does $x \to x_0$ always imply $f(x) \to L$? You have me confused. Commented Mar 16, 2021 at 3:28
• @KraZZ the point is that in the definition of $f(x)\to L$ as $x\to x_0$ we generally take the extra condition that $x$ is not allowed to actually equal $x_0$ (as in the definition you give above). Commented Mar 16, 2021 at 8:57