- #1
alyafey22
Gold Member
MHB
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Let $f: \mathbb{R}\to \mathbb{R}$ be such that $f(x+y)=f(x)+f(y)$ for all $x,y \in \mathbb{R}$ . Assume the limit of $\lim_{x\to 0}f(x)=L$ then prove that $L=0$.
Here is my attempt:
We first choose an arbitrary \(\displaystyle \epsilon >0\) then we choose $\delta$ such that \(\displaystyle \text{Max}(|x|,|y|,|x+y|)<\delta \) so we can say that \(\displaystyle |f(y)-L|<\frac{\epsilon}{2} \) and \(\displaystyle |f(x+y)-L|<\frac{\epsilon}{2} \)
Since we have the following
\(\displaystyle |f(x+y)-L|=|f(x)+f(y)-L| \geq |f(x)|-|f(y)-L|\)
\(\displaystyle |f(x)|\leq |f(x+y)-L|+|f(y)-L|<\frac{\epsilon}{2} +\frac{\epsilon}{2} =\epsilon \) for the chosen \(\displaystyle \delta \) by existence of limit of \(\displaystyle f\).
So we deduce that
\(\displaystyle 0<|x|<\delta \) implies \(\displaystyle |f(x)|<\epsilon \)
since \(\displaystyle \epsilon >0\) is arbitrary we have \(\displaystyle \lim_{x \to 0}f(x)=0 \,\,\, \square \).
So what you think guys ?
Here is my attempt:
We first choose an arbitrary \(\displaystyle \epsilon >0\) then we choose $\delta$ such that \(\displaystyle \text{Max}(|x|,|y|,|x+y|)<\delta \) so we can say that \(\displaystyle |f(y)-L|<\frac{\epsilon}{2} \) and \(\displaystyle |f(x+y)-L|<\frac{\epsilon}{2} \)
Since we have the following
\(\displaystyle |f(x+y)-L|=|f(x)+f(y)-L| \geq |f(x)|-|f(y)-L|\)
\(\displaystyle |f(x)|\leq |f(x+y)-L|+|f(y)-L|<\frac{\epsilon}{2} +\frac{\epsilon}{2} =\epsilon \) for the chosen \(\displaystyle \delta \) by existence of limit of \(\displaystyle f\).
So we deduce that
\(\displaystyle 0<|x|<\delta \) implies \(\displaystyle |f(x)|<\epsilon \)
since \(\displaystyle \epsilon >0\) is arbitrary we have \(\displaystyle \lim_{x \to 0}f(x)=0 \,\,\, \square \).
So what you think guys ?