| Line 28: | Line 28: | ||
Indeed, let us select <math> \varepsilon >0 </math>. Since (*) is true, then | Indeed, let us select <math> \varepsilon >0 </math>. Since (*) is true, then | ||
<math>\exists \delta(\varepsilon)>0 </math> such that | <math>\exists \delta(\varepsilon)>0 </math> such that | ||
| − | <math> 0 < |z-c| < \delta </math> implies <math> |f(z)-L|< \varepsilon </math>. | + | <math> 0 < |z-c| < \delta \, </math> implies <math> |f(z)-L|< \varepsilon </math>. |
Consider <math> |h(x)-L| \, </math> when <math>x\to 0 </math>. | Consider <math> |h(x)-L| \, </math> when <math>x\to 0 </math>. | ||
Revision as of 09:23, 7 April 2010
To ask a new question, add a line and type in your question. You can use LaTeX to type math. Here is a link to a short LaTeX tutorial.
To answer a question, open the page for editing and start typing below the question...
go back to the Discussion Page
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(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 $
