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The Precise Definition of a Limit
Previously we stated that intuitively the notion of a limit is the value a function approaches at a given point. We refined this notion in terms of approximations, stating that
lim x→x 0 f (x) = L
if for any error tolerance around L, we can find an interval around x 0 (excluding x 0 ) such that for all x values within that interval, the distance between f (x) and L is within the error tolerance. In order to prove mathematically that a limit does or does not exist, it is much more useful to cast this definition in the language of mathematics. Without further ado, the formal definition of a limit.
Definition: The Limit Suppose f (x) is defined on an open interval about x 0 , not necessarily containing x 0. We say that L is the limit of f (x) as x approaches x 0 , written
lim x→x 0 f (x) = L
if for every number > 0, there exists a corresponding number δ > 0 such that for all x with 0 < |x − x 0 | < δ we have
|f (x) − L| < .
This definition is the same as the definition we stated above, but we have rewritten some of the verbal terms such as ‘error tolerance’ in terms of variables. We can describe the math- ematical definition in words as follows: for any error tolerance ( > 0), we can restrict the values of f (x) to be within that error tolerance of L (|f (x) − L| < ) by restricting the values of x to be sufficiently close but not equal to x 0 (by choosing δ > 0 and considering x with 0 < |x − x 0 | < δ). We require 0 < |x − x 0 | in order to exclude the point x 0 from the interval of interest, because the limit does not depend on the function value at that single point. It is also noteworthy that when we use the phrase ‘for every > 0,’ it is sufficient to show that it is true for an arbitrary (because we could just as well replace ‘this’ with any and ‘every’ number and the analysis would still hold true). We have recasted the verbal definition so that we can manipulate the variables and δ in order to make precise calculations to prove whether or not a limit exists. Let us consider the following examples in order to further understand this definition.
Example 1 Given f (x) = 2x + 1
is it possible to find a δ-interval around x 0 = 2 so that for x with 0 < |x − 2 | < δ it follows that |f (x) − 5 | <
if = 0.5? Using this information can you conclude whether or not
lim x→ 2
f (x) = 5
is a true statement? Solution We will begin with the inequality we want to satisfy, and try to deduce a restriction on x in order to satisfy it.
|f (x) − 5 | < | 2 x + 1 − 5 | < 0. 5 | 2 x − 4 | < 0. 5 2 |x − 2 | < 0. 5 |x − 2 | < 0. 25
Since the inequality |f (x) − 5 | < is true for |x − 2 | < 0 .25, it follows that δ = 0.25 will define an interval sufficiently small. Note that any smaller value for δ will also work, so we could just as well let δ = 0.1. Either way, we conclude that it is possible to find such a δ-interval. Although lim x→ 2 f (x) = 5
is in fact a true statement, being able to find δ for a single error tolerance (as we have just done) is insufficient information to make this conclusion (we would need to find the width of the δ-interval for every .)
Example 2 Given
f (x) = sin(
x
is it possible to find a δ-interval around x 0 = 0 so that for x with 0 < |x| < δ it follows that
|f (x) − 0 | <
if = 0.1? Using this information can you conclude whether or not
lim x→ 0 f (x) = 0
is a true statement? Solution We will need to approach this problem differently, because there is no way for us to algebraically manipulate the inequality
| sin(
x
in order to reach a restriction for x. Instead, let us consider the problem graphically. We see that the function values are oscillating between -1 and 1, and the oscillations become more rapid as x → 0. Because these oscillations continue indefinitely close to x = 0, no matter
Example 3 Prove that lim x→ 1 f (x) = 3 if f (x) = 2x + 1.
Solution Consider > 0, arbitrary. We must find a corresponding δ > 0 such that for 0 < |x − 1 | < δ we have |f (x) − 3 | < . Our strategy will be to analyze the second inequality to try to determine δ.
|f (x) − 3 | = |(2x + 1) − 3 | = | 2 x − 2 | = 2|x − 1 |
and 2 |x − 1 | < if |x − 1 | <
Thus, if we let δ =
we find that for arbitrary x with 0 < |x − 1 | < δ it follows
|f (x) − 3 | = 2|x − 1 | < 2 ·
From the definition of the limit, it follows that
lim x→ 1 f (x) = 3
Example 4 Prove that lim x→ 3 f (x) = 9 for
f (x) =
x^2 x 6 = 3 3 x = 3
Solution Consider > 0, arbitrary. We must find a corresponding δ > 0 such that for 0 < |x − 3 | < δ we have |f (x) − 9 | < . Since 0 < |x − 3 |, it follows x 6 = 3, so f (x) = x^2. Thus, |f (x) − 9 | = |x^2 − 9 | = |(x + 3) · (x − 3)| = |x + 3| · |x − 3 |
We can control the value of |x − 3 | directly with δ, but we will have to control |x + 3| indirectly. If we knew that δ would be less than 1, then we’d know from 0 < |x − 3 | < δ that 3 < x < 5. Now the expression
|f (x) − 9 | = |x + 3| · |x − 3 | <
consists of two terms we can control. If δ < 1, then the maximum value of x is 5, so the max- imum value of |x + 3| is 8. It follows, granted δ < 1, that if |x − 3 | < /8, then |f (x) − 9 | < . Thus, choose δ = min(1, /8).
Consider arbitrary x with 0 < |x − 3 | < δ. For such an x, it follows
|f (x) − 9 | = |x + 3| · |x − 3 | < 8 ·
Example 5 Prove that lim x→ 0
f (x) 6 = 3 for
f (x) =
x
Solution Unlike the previous examples, where we needed to find a δ-interval that forces the values of f (x) to be sufficiently close to L for an arbitrary , our task here is to find a single value for for which it is impossible to construct any δ-interval which restricts the function to be within of L = 3 (in other words, for any δ-interval we need to demonstrate at least one value x for which |f (x) − L| > ). Because this particular function f (x) grows without bound as x → 0 from the right and decreases without bound as x → 0 from the left, we can definitely find such a value for ; it is just a matter of coming up with one. One method would be to simply guess a very small value for , such as = 10−^10 and check if it will work. In fact, we could in theory choose any finite value for in this case, but let us just use a simple number such as = 1 (the mechanics of this problem only become a bit more difficult if we choose > 3, because of the way the function is separated around the y-axis).
Now we consider δ > 0, arbitrary. If we let
x = −
δ 2
we will have
0 < |x − 0 | =
δ 2
< δ
yet
|f (x) − 3 | = |
−δ/ 2
δ
Thus, we have just proven that
lim x→ 0
x
It is noteworthy that we could just as well replace 3 with any positive, finite value L and the same type of analysis would work. If we wanted to consider a negative value for L, we would just use
x = −
δ 2
instead. Still, if we wanted to consider L values small in magnitude, we would correspond- ingly need to choose smaller values for in order to show the limit does not exist (at least in this way). We are able to perform such analysis for any L value with ease because f (x) is positive for positive x and negative for negative x, which conveniently separates this function around the y-axis.
Example 6 Prove that lim x→ 0 f (x) 6 = 10 for
f (x) =
|x|
Solution Once again we need to find a value for so that for any δ > 0 we can find an x with 0 < |x − 0 | < δ and |f (x) − 10 | > . Looking at this function graphically we can once again see this must be possible, but because we do not have the same separation of positive and negative values for f (x) as we cross the y-axis, the mechanics of this problem are a little
