if $f(x)=g(x)$ when $|x-a|<\lambda$.prove that $\lim_{x \to a} f(x)=\lim_{x \to a} g(x)$

Question

suppose there is a $\lambda>0$ such that $f(x)=g(x)$ when $|x-a|<\lambda$.prove that $lim_{x \to a} f(x)=lim_{x \to a} g(x)$

my attempt:

assume that :$lim_{x \to a} f(x)=l$

that's means $\forall \epsilon$ there is some $\lambda'$ such that if $0<|x-a|<\lambda'$ then $|f(x)-l|<\frac{\epsilon}{2}$

now let's assume $lim_{x \to a} g(x)=m$

and $l \neq m$

so that's also means:

that's means $\forall \epsilon$ there is some $\lambda''$ such that if $0<|x-a|<\lambda''$ then $|g(x)-m|<\frac{\epsilon}{2}$

now if $|x-a|<\lambda=min(\lambda ',\lambda '')$ then obviously $|f(x)-g(x) +m-l|<\epsilon$ ,and with this $g(x)=f(x)$ we can write the last statement as $|m-l|<\epsilon$, wich is not true for all $\epsilon >0$ (take $\epsilon = \frac{|m-l|}{2}$),so our assumption ($l \neq m$) is of course false,that's means $l=m$,so finally if there is some $\lambda $ such that if $|x-a|<\lambda$ then $f(x)=g(x)$,then $lim_{x \to a} g(x)=lim_{x \to a} f(x)$ .

my question :-does my proof is right?

(if you know other method you can post it ,but first you should answer to my question )

Answer 1

1. As someone else said in the comment section, we don’t know if the limits exist without knowing anything about the functions f and g (f and g can be infinity at a). To make your proof work you need to specify some properties of f and g.

2. If f and g are both defined at a, you can simply say that the limits equal f(a) and g(a); by hypothesis you automatically have lim f = lim g.

Answer 2

If the limit of f exists at x=a then choosing $\delta=\lambda$ in the definition of limit leads to the existence of the limit of g, since g = f for a vicinity no farther than $\lambda$ from value a.

Answer 3

In your attempt you have to assume that $\displaystyle\lim_{x\to a}g(x)$ exists. In my opinion, this is imprecise.

The right statement is: Let $X\subset\mathbb R$, $a\in X'$ and $f,g\colon X\to\mathbb R$ be given. If $\displaystyle\lim_{x\to a}f(x)=L$ and there is $\delta_0>0$ such that $f(x)=g(x)$ for each $x\in X$ with $0<|x-a|<\delta_0$, then $\displaystyle\lim_{x\to a}g(x)=L$. Now, I prove the statement.

Let $\varepsilon>0$ be given. By definition of limit, there is $\delta_{\varepsilon}'>0$ such that if $x\in X$ and $0<|x-a|<\delta_\varepsilon'$, then $|f(x)-L|<\varepsilon$. So, if $\delta_\varepsilon=\min\{\delta_0,\delta_\varepsilon\}$ and $x\in X$ satisfies $0<|x-a|<\delta_\varepsilon$, we have $f(x)=g(x)$, and so \begin{gather*} |g(x)-L|=|f(x)-L|<\varepsilon. \end{gather*} Thus $\displaystyle\lim_{x\to a}g(x)=L$.