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- More on Limits When we gave the formal definition of a limit we assumed that a and L were finite. However, there are occasions when this point of view is too limiting. We state here some limit definitions when either a or L (or both) is infinite.
Definition 1. (i) Let I be an open interval containing a. Let f be a (real-valued) function defined on I except possibly at a. We say that f approaches ∞ as x approaches a if given N > 0 there exists δ > 0 such that f (x) > N whenever 0 < |x − a| < δ. In this situation, we write lim x→a f (x) = ∞. We say that f approaches −∞ as x approaches a if given N > 0 there exists δ > 0 such that f (x) < −N whenever 0 < |x − a| < δ. In this situation, we write lim x→a f (x) = −∞. With some minor modifications, this definition can be modified to one-sided limits. (ii) Let I be an interval of the form (b, ∞). Let f be a function defined on I. We say that f approaches L (a finite number) as x approaches ∞ if given > 0 there exists M > 0 such that |f (x) − L| < whenever x > M. In this situation, we write lim x→∞ f (x) = L. We say that f approaches ∞ as x approaches ∞ if given N > 0 there exists M > 0 such that f (x) > N whenever x > M. In this situation, we write lim x→∞ f (x) = L. We say that f approaches −∞ as x approaches ∞ if given N > 0 there exists M > 0 such that f (x) < −N whenever x > M. In this situation, we write lim x→∞ f (x) = L. Limits as x approaches −∞ are defined analogously.
Example 2. Let f (x) = (^) x^12. Then lim x→ 0 f (x) = 0.
Proof. Let N > 0. (When computing limits rigorously this is an excellent first statement.) We need to find δ > 0 such that if 0 < |x| < δ then f (x) > N. Now if |x| < δ then (^) |x^1 | > (^1) δ and hence (^) |x^1 | 2 > (^) δ^12. By choosing δ > 0 so that
δ^2
= N,
the desired condition on δ will be satisfied. Solving for δ in equation 1 we obtain
δ =
N
Thus, if 0 < |x| <
N
then f (x) > N as required. 1
The above proof contained a lot of repetition. As one acquires a facility with these kinds of proofs much of the repetition can be dispensed with. However, there is rarely any harm in overwriting a proof, and you are encouraged to write a lot when asked to provide a proof. In any case, you will only occasionally be asked to give formal proofs since the focus of the calculus course here at Weber State is computational. There is a time and place for serious proof writing and it is called analysis. (Math 3810, 4210, and 4220 at Weber State.)
Example 3. Let f (x) = sinx^ x. Then lim x→∞ f (x) = 0.
Proof. Let > 0. We need to find M > 0 such that |f (x)| < whenever x > M. Now if x > M > 0
then 1 M
x
Since | sin x| ≤ 1 for any real number x, it follows that
(2) |f (x)| =
sin x x
x
M
By choosing M = (^1) the calculation in (2) yields the desired result.
There are several important limit laws that can aid with calculations. Some of the laws are mentioned in the text. We mention a few more here.
Theorem 4. Suppose that a is a finite number, and f is a function defined on an interval containing a except (possibly) at a. Then lim x→a exists if and only if both lim x→a+
and lim x→a−
exist and are equal.
This result is frequently used to show that a limit does not exist. For example, lim x→ 1
x − 1
does not exist because the two one sided limits are not equal.
Theorem 5. In this result, a may be finite or ±∞. Suppose that lim x→a f (x) = ∞, lim x→a g(x) = ∞, and lim x→a h(x) = c where f , g, and h are functions defined near a
and c is a positive number. Then the following hold: (i) lim x→a (f (x) + g(x)) = ∞ (ii) lim x→a f (x)g(x) = ∞ (iii) lim x→a f (x)h(x) = ∞ (iv) lim x→a f (x)(−g(x)) = −∞ (v) lim x→a f (x)(−h(x)) = −∞
Notice that there was no result for lim x→a (f (x) − g(x)). This is because that limit
may not exist, and even when it does, the value of the limit could be anywhere in [−∞, ∞]- that is, any real number or ±∞. The expression
lim x→a f (x) − lim x→a g(x) = ∞ − ∞
is an example of an indeterminate form. Another indeterminate form is 0 · ∞. We will discuss such forms in the second semester of this course when we look at l’Hospital’s rule.
If f is not continuous at a then f is discontinuous at a.
For technical reasons, it is often convenient for us to define continuity on closed intervals rather than open.
Theorem 11. If f and g are continuous at a and c is a constant, then f ± g, f g, and cf are all continuous at a. If g(a) 6 = 0, then fg is also continuous at a.
The proof follows immediately from the corresponding results for limits.
Theorem 12. If g is continuous at a and f is continuous at g(a) then f ◦ g is continuous at a.
Thus, the composition of continuous functions is continuous.
Example 13. Let P, Q be polynomials with Q not identically 0. Then P is con- tinuous on all of R, and PQ is continuous away from the roots of Q. Trigonometric functions and root functions are also continuous on their domains. In previous courses, you may have seen exponential and logarithmic functions. These functions are also continuous.
It is often worthwhile considering how a function can fail to be continuous.
Definition 14. Suppose that lim x→a+
f (x) = L and lim x→a−
f (x) = M with L and M
both finite. If L = M but f (a) 6 = L (possibly because f (a) is undefined) then f has a removable discontinuity at x = a. If f is undefined at x = a and a is a removable discontinuity of f at x = a, a continuous extension of f to a is given by
(3) g(x) =
f (x) x 6 = a lim x→a f (x) x = a
Example 15. Note that f (x) = x
2 x is undefined at^ x^ = 0. However, the disconti- nuity is removable, and g(x) = x is a continuous extension of f to x = 0.
Definition 16. Suppose that lim x→a+^
f (x) = L and lim x→a−^
f (x) = M with L and M
both finite. If L 6 = M , then f has a jump discontinuity at a.
Example 17. The discontinuities of the greatest integer function are jump discon- tinuities.
Definition 18. If f is discontinuous at a but the discontinuity is neither a jump discontinuity nor a removable discontinuity then the discontinuity is essential. If at least one of the one-sided limits is ±∞ then the discontinuity is infinite.
Example 19. The function f (x) = sin
1 x
has an essential discontinuity at x = 0.
(Try to graph f on a graphing calculator or on a computer algebra system.)
Example 20. If P and Q are non-zero polynomials, and P (a) 6 = 0 = Q(a) then PQ has an infinite discontinuity at x = a.
Exercise 21. Classify the discontinuities of each of the following functions:
(a) f (x) =
x x^6 = 0 3 x = 0
(b) f (x) = (x+1)
3 x^2 − 1 (c) f (x) = x sin
1 x
x 6 = 0
(d) f (x) =
4 − x^2 x ≥ 0 x^3 x < 0
Exercise 22. Explain why
f (x) =
x^2 cos
1 x^4
x 6 = 0 0 x = 0
is continuous on all of R.
Definition 23. Suppose that f is a function defined on a closed bounded interval I. If f has finitely many discontinuities, none of which are essential, then f is piecewise continuous on I.
Example 24. Any continuous function on I is piecewise continuous. The greatest integer function is piecewise continuous on any closed bounded interval. The func- tion f (x) = (^) x 2 x− 4 fails to be piecewise continuous on any closed bounded interval containing 2 or −2.
Piecewise continuity plays an important role when we think about integration. We close with an important result relating differentiability and continuity.
Theorem 25. If f is differentiable at a then f is continuous at a.
Proof. We need to show that lim x→a f (x) = f (a).
Exercise 26. a) For every real number x there is an integer n (depending on x) such that n ≤ x < n + 1. The number n is called the greatest integer of x and is denoted by n = [x]. Sketch the graph of y = [x] for − 5 ≤ x ≤ 5. Where is [x] continuous? Where is [x] differentiable? If y = [x] is differentiable at a compute dydx at x = a. b) It can be shown that lim θ→ 0 cos θ = 1 and that lim θ→ 0 sin θ = 0. Hence, sin and cos
are continuous at θ = 0. Use the sum rules for sine and cosine to show that sin and cos are continuous on the entire real line.
