- Thread starter
- #1

- Jan 17, 2013

- 1,667

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 ?