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 July 5th, 2019, 12:31 PM #1 Senior Member   Joined: Oct 2015 From: Greece Posts: 137 Thanks: 8 I can't understand how to prove limits using the definition... $\displaystyle \lim_{x \to c} f(x) = L$ if $\displaystyle \forall ε > 0$, $\displaystyle \exists δ>0$ : $\displaystyle 0 < |x-c| < δ$ => $\displaystyle |f(x) - L| < ε$ , $\displaystyle \forall x$ maybe except c. First of all do I understand what the definitions says? My understanding is this: "The closer x is at c the more precisely f(x) will approach L." In order to define this mathematically we need two things. An number δ to precisely specify the words "The closer x is at c" and another number ε which precisely describes the words "the more precisely f(x) will approach L". If ε is really small, the more precise the limit will be. Some exercises which I'm trying to solve: Exercise 1: 1.jpg Exercise 1 (continues): 2.jpg Exercise 2: 3.jpg For example in exercise 1 he finds out that: $\displaystyle |x-1| < ε/5$ Then i can not see why he chooses δ=ε/5 . Is it because both in-equations are similar? $\displaystyle |x-1| < ε/5$ $\displaystyle 0 < |x-1| < δ$ so it seems like δ = ε/5 if you compare them. Is there a more mathematically procedure to actually show that? Can you explain the solution of ex1 better that the book? Thank you.
 July 6th, 2019, 04:19 AM #2 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,683 Thanks: 2662 Math Focus: Mainly analysis and algebra The limit definition says that [B]given a value for $\epsilon$, we can always pick a value for $\delta$ (the maximum difference from the limit point) such that the function is within $\epsilon$ of the limit. So, given any $\epsilon$ we find that $|x-1| < \frac{\epsilon}{5}$. So whatever $\epsilon$ may be, if we pick our $\delta$ to be $\frac{\epsilon}{5}$, we will have $|x-1| < \frac{\epsilon}{5} = \delta$ which, since $a < b = c \implies a < c$ gives us the $|x-1| < \delta$ which we require. All of that is preparatory work. What follows is the proof that picking $\delta = \frac{\epsilon}{5}$ delivers the required inequality $|f(x)-2| < \epsilon$. Thanks from romsek and babaliaris
 July 6th, 2019, 03:08 PM #3 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,683 Thanks: 2662 Math Focus: Mainly analysis and algebra It's worth pointing out that, although we found (in the first part) an inequality that suggested a value for $\delta$, the method doesn't always indicate that there are no other places where $f(x)$ is close to $L$. This is why we need to do the second part, which is the actual proof. Also, since $|x-1| < \frac{\epsilon}{5}$ describes a symmetrical interval about $x=1$ (for any given value of $\epsilon$), it should be clear that we could instead pick a smaller value for $\delta$ if we so desired, $\frac{\epsilon}{10}$ for example. Thanks from babaliaris

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