# A different way to prove $\lim\limits_{x\to x_0}g(x)$ with $\epsilon,\delta$

for $f,g: \mathbb R\to \mathbb R\;\;and\;\; x_0\in \mathbb R$

Assume that $$\lim\limits_{x\to x_0}f(x)=L$$

prove: for every $\epsilon>0$ there is $\delta>0$ so that for every x that sustains $\delta>|x-x_0|>0$ and $\epsilon> |f(x)-g(x)|$ so there is$\lim\limits_{x\to x_0}g(x)=L$

i dont know even how to approach it, thanks in advance for the help

Let $\epsilon > 0$ be arbitrary. Then $\exists \delta_1 > 0$ such that

$$0<|x-x_0|<\delta_1 \Rightarrow |f(x_0) - L| < \epsilon/2$$

and $\exists \delta_2 > 0$ such that

$$0<|x-x_0|<\delta_2 \Rightarrow |f(x) - g(x)| < \epsilon/2.$$

Put $\delta := \min\{\delta_1,\delta_2\}$. Then by the triangle inequality,

$$0<|x-x_0|<\delta \Rightarrow |g(x) - L| \leq |f(x) - L| + |f(x) - g(x)| < \epsilon/2 + \epsilon/2 = \epsilon.$$

NB I hope I have interpreted your question correctly.

• exactly! thanks. i got to $2\epsilon>|f(x)-g(x)|$ but now i understand everything – user137645 Mar 24 '14 at 16:58