Let $a \in \mathbb{R}$ and $I$ be an open interval with $a \in I$. Let $f$ and $g$ be real functions defined on $I$ except possibly at $a$. If $f(x) = g(x) \forall x \in I \backslash \{a\}$ and $\lim_{x \rightarrow a} f(x) = L$ prove that $g(x) \rightarrow L$ as $x \rightarrow a$.

So far what i have is:

Aim: Get a contradiction when I assume $lim_{x \rightarrow a} g(x) \neq L$.

Since $lim_{x \rightarrow a} g(x) \neq L$ then $\exists \epsilon \gt 0 \forall \delta \gt 0$ \exists x\in I\blackslash {a} $0<\lvert x - a\rvert \lt \delta $ and $ \lvert g(x) - L\rvert \geq \epsilon$. Since $g(x) = f(x)$ then $\lvert f(x) - L\rvert \geq \epsilon$. This is in contradiction with

$lim_{x\to a}f(x)=L$.

Are my steps so far correct and if they are could someone tell me how to proceed next as I am stuck here.


Hint. Yes, you have to use the definition of limit $\lim_{x \rightarrow a} f(x) = L$, but your proof is not correct (we do not know anything about $g(a)$ or $f(a)$).

Now note that in such definition we never use the value of $f$ at $a$.


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