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| − | |f(x+c)-L|=|h(x)-L|<\varepsilon | + | |f(z)-L|=|f(x+c)-L|=|h(x)-L|<\varepsilon |
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Latest revision as of 09:27, 7 April 2010
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Could someone formulate 4.1.4? It's easy but I'm not sure my answer is correct. Thanks
prof. Alekseenko: In problem 4 we need to prove the "if and only if" statement. It has to parts. The first part is that
if $ \lim_{x\to c} f(x) = L $ then $ \lim_{x\to 0} f(x+c) = L $.
The second part goes in the opposite direction. Namely, we need to prove that
if $ \lim_{x\to 0} f(x+c) = L $ then $ \lim_{x\to c} f(x) = L $.
Let me work out the first part and may be somebody can do the second. We assume that
(*) $ \lim_{z\to c} f(z) = L $
Notice that we used letter $ z\, $ instead of the variable $ x\, $. The reason for this will be clear soon. However, using a different letter to denote the variable is not a problem.
Let us show that $ h(x):=f(x+c)\, $ also has limit at $ x=0\, $ and that this limit is $ L \, $.
Indeed, let us select $ \varepsilon >0 \ $. Since (*) is true, then $ \exists \delta(\varepsilon)>0 \, $ such that $ 0 < |z-c| < \delta \, $ implies $ |f(z)-L|< \varepsilon $.
Consider $ |h(x)-L| \, $ when $ x\to 0 \, $. Notice that if $ |x-0|<\delta \, $. If $ z=x+c \, $ then
$ |x-0|=|(x+c) - c|= |z-c|<\delta \, $
Then, of course,
$ |f(z)-L|<\varepsilon $
However, $ z=x+c \, $, therefore,
$ |f(z)-L|=|f(x+c)-L|=|h(x)-L|<\varepsilon $
In other words, $ |h(x)-L|<\varepsilon $ as long as $ |x-0|<\delta\, $. Because $ \varepsilon>0 $ was arbitrary, we conclude that
(**) $ \lim_{x\to 0} h(x) = \lim_{x\to 0} f(x+c) = L $
