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$$f;g:[a,b] \rightarrow R \, \, \forall x \in (a;b) \, \, g'(x) \neq 0,$$ $f,g$ are differentiable in $(a,b)$. If $$\lim_{x \to a} f(x) = \lim_{x \to a }g(x) = 0$$ and there exists a $$\lim_{x \to a} \frac{f'(x)}{g'(x)},$$ then $$\lim_{x \to a} \frac{f(x)}{g(x)}=\lim_{x \to a} \frac{f'(x)}{g'(x)}$$

Then two functions are constructed $F(x) =\begin{cases} f(x),&\text{if }x\neq a\\0,&\text{if }x=a\end{cases}$ and $G(x) =\begin{cases}g(x),&\text{if }x\neq a\\0,&\text{if }x=a\end{cases}$.

My question is about continuity of the functions $F'(x), G'(x)$. They both are continuous near point $a$, except when $x=a$. In the proof it's explained that they are also continuous at $x=a$, because at $x=a$, $F(x)=\lim_{x \to a}f(x) = 0$. I don't get why it's possible to express $F(x)$ as $\lim_{x \to a}f(x)$ when $x= a$. Or rather why the fact that $x\neq a$ implies that $x$ tends to $a$. Can someone explain?

Simply Beautiful Art
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user
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2 Answers2

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It is not that it is possible to express...etc., it is just that that's the way we have to check continuity of $\;F\;$ at $\;a\;$ . In fact, we have that

$$F(x)=\begin{cases}f(x),&x\neq a\\{}\\0,&x=a\end{cases}\implies\;F\;\text{ is continuous at }\;a\iff \lim_{x\to a}F(x)=F(a)=0$$

BTW, you also need for L'H rule the condition $\;g'(x)\neq 0\;$ in some neighborhood of $\;a\;$ ...

DonAntonio
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You have an assumption that $f $ has a limit $0$ at $a$, so when you define $F $ to be equal $f$ near $a$ and equal a limit of $f$ at $a$ then $F$ is continuous, because its value at $a$ equals its limit at $a$.
Same for $G$.

CiaPan
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