Precise Definition of a Limit
In the previous sections, we discussed limits intuitively. However, the precise definition of a limit can be made more detailed. See the following definition.
Definition:
Let f be a function defined on an open interval that contains a, except at a itself. The limit of f(x) as x approaches a is L:
If for every number \epsilon > 0 there is a number \delta > 0 such that:
If 0 < |x-a| < \delta then |f(x) – L| < \epsilon
Let’s walk through an example to illustrate the precise definition of a limit concept. Take a look at the function below, which is also graphed as shown.
f(x) = \left\{\begin{matrix}2x-1, &x\neq 3&\\6,\space\space& x = 3&\end{matrix}\right.
We can see that when x is close to 3 but not equal to 3, then f(x) is close to 5. This means that lim_{x \rightarrow 3}(2x-1) = 5.
Well, how can we instead use the precise definition of a limit to prove this? The first step is to identify that a = 3 and L = 5. First, let’s say that we want to find how close to 3 does x need to be so that the difference between 5 and f(x) is less than 0.1.
So, from the precise definition of the limit, we need to find a number \delta > 0 so that the following is true:
|f(x) – L| < \epsilon when 0 < |x-a| < \delta
or
|f(x) – 5| < 0.1 when 0 < |x-3| < \delta
Continuing along, notice that if 0 < |x-3| < \frac{0.1}{2} = 0.05 then:
|f(x) – 5| = |(2x-1) – 5| = |2x-6| = 2|x-3| < 0.1, so:
|f(x) – 5| < 0.1 when 0 < |x-3| < 0.05
This means that if x is within a distance of 0.05 from 3, then the value of f(x) will be within a distance of 0.1 from 5.
Now, if we change the number 0.1 to a smaller number such as 0.01 or 0.001, we have:
|f(x) – 5| < 0.01 when 0 < |x-3| < 0.005
or
|f(x) – 5| < 0.001 when 0 < |x-3| < 0.0005
Based on this, we find that:
|f(x) – 5| < \epsilon when 0 < |x-3| < \delta = \frac{\epsilon}{2}
This is essentially the precise way of saying “the value of f(x) is near 5 when x is near 3. The figure illustrates what we have just discussed.
By taking any value of x that lies in between 3 – \delta and 3 + \delta, we can make the values of f(x) lie in between 5 – \epsilon and 5+\epsilon
Example 6
Using the precise definition a limit, prove the following:
lim_{x \rightarrow 0}(x^2) = 0
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