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**emakarov** One has to show:

for every $\displaystyle \varepsilon>0$ there exists a $\displaystyle \delta>0$ such that for all $\displaystyle x\in X$, $\displaystyle 0<|x-a|<\delta$ implies $\displaystyle |f(x) - L| < \varepsilon/2$. (*)

Since $\displaystyle |f(x) - L| < \varepsilon/2$ is equivalent to $\displaystyle |2f(x) - 2L| < \varepsilon$, (*) means that $\displaystyle \lim\limits_{x\to a}2f(x)=2L$.

How do we show (*)? We are given

for every $\displaystyle \varepsilon>0$ there exists a $\displaystyle \delta>0$ such that for all $\displaystyle x\in X$, $\displaystyle 0<|x-a|<\delta$ implies $\displaystyle |f(x) - L| < \varepsilon$. (**)

The latter statement means that there is a jinny that, given any $\displaystyle \varepsilon$, returns a $\displaystyle \delta$ such that for every x bla, bla, bla. Now, you need to do something similar to that jinny to show (*). Suppose somebody gives you an $\displaystyle \varepsilon$. You halve it and give $\displaystyle \varepsilon/2$ to the jinny. True to its task, the jinny gives you back a $\displaystyle \delta$ such that for all $\displaystyle x\in X$, $\displaystyle 0<|x-a|<\delta$ implies $\displaystyle |f(x) - L| < \varepsilon/2$. You take this $\displaystyle \delta$ and return it as your own.