From math^2

Calculus: DefinitionOfALimit

In the previous article, we said that the limit of f as x approaches to a number a is L. This only means that the value of f can be made closer to L as we please by taking x close enough to a. Since we want f very close to L at a, we make the absolute difference of f(x) and L (|f(x)-L|) very small by making the absolute difference of x and a (|x-a|) small enough. Let's notate this "smallness" by two symbols epsilon (\epsilon) and delta (\delta).

Epsilon (\epsilon) and delta (\delta) in limits..

Epsilon and delta are two Greek symbols which are widely used in mathematics and sciences. The convention meaning for epsilon is "small positive infinitesimal quantity" while delta is "difference" or "a change". In this case, the value of \delta is dependent on \epsilon. Therefore, if the limit of the function f(x) as x approaches a is L, then for every positive number \epsilon , no matter how small it is, there exists a corresponding positive number \delta such that |f(x)-L< \epsilon| whenever 0<|x-a|<\delta. This leads to the formal definition of the limit concept:

Formal Definition of a Limit

Let f be a function defined at all point x on the open interval I that contains a, except possibly at a itself. Then, the limit of the function f(x) as x approaches a is L, which is notated as,

\lim_{x \to a}f(x)=L

if for every \epsilon > 0 there exists a \delta > 0 such that |f(x) - L | < \epsilon whenever 0 < |x-a| < \delta. Now, using the definition, we will prove that the following function has its limit as indicated.

Example #1

Using the formal definition of a limit discussed above, verify that

\lim_{x \to 5}(2x-7)=3.

According to the definition, we have to show that for every \epsilon > 0 there exists a \delta > 0 such that <? |(2x-7)-3|< \epsilon whenever 0<|x-5|> \delta.



Thus, accordingly, it must be shown that

2|x-5|< \epsilon whenever 0<|x-5|> \delta

which can also be written as

|x-5|< \frac{\epsilon}{2} whenever 0<|x-5|> \delta.

Look at equation (5), seems we have our perfect choice for \delta. Proceeding accordingly, we choose \delta = \frac{1}{2} \epsilon and obtain

2|x-5|< 2 \delta whenever 0<|x-5|> \delta \end{equation}

which is just equal to

|(2x-7)-3|< \epsilon whenever 0<|x-5|> \delta

Thus, as clearly shown and defined,

\lim_{x \to 5}(2x-7)=3

We have discussed the definition of a limit and directly applied it in finding limits. With this information alone, we can solve and evaluate any given limits. However, solving using the methods employed above can be very tedious. In the previous article on "limits of a function", we looked at smarter methods and techniques to find limits with the help of several limit theorems.

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Page last modified on June 26, 2011