Premed · Premed · Calculus 1
Lecture 3: Computing Limits and Continuity
Calculus I — Differential Calculus
Learning Objectives
By the end of this lecture, students will be able to:
- Apply the limit laws to evaluate limits algebraically
- Use algebraic techniques (factoring, rationalizing, common denominators) to resolve indeterminate forms
- Apply the Squeeze Theorem to evaluate limits
- Define continuity at a point and on an interval
- Identify types of discontinuities
- Apply the Intermediate Value Theorem
Lecture Content
I. Limit Laws
Suppose lim_{x -> a} f(x) = L and lim_{x -> a} g(x) = M. The limit laws provide a toolkit for computing limits algebraically. The Sum Law states that lim_{x -> a} [f(x) + g(x)] = L + M, and the Difference Law gives lim_{x -> a} [f(x) - g(x)] = L - M. The Constant Multiple Law says lim_{x -> a} [c f(x)] = c L. The Product Law yields lim_{x -> a} [f(x) g(x)] = L M. The Quotient Law tells us lim_{x -> a} [f(x)/g(x)] = L/M, provided M is not zero. The Power Law gives lim_{x -> a} [f(x)]^n = L^n for any positive integer n, and the Root Law extends this to fractional powers under appropriate conditions.
A particularly useful consequence of these laws is the direct substitution property: if f is a polynomial or rational function and a lies in the domain of f, then lim_{x -> a} f(x) = f(a). In other words, for these well-behaved functions, computing the limit is as simple as plugging in the value.
II. Evaluating Limits Algebraically
When direct substitution fails, several algebraic strategies can resolve the issue. The first step is always to try direct substitution. If it yields a real number, that is the limit. If it yields the indeterminate form 0/0, further work is required.
Factoring and canceling is often the simplest approach. For example, lim_{x -> 2} (x^2 - 4)/(x - 2) gives 0/0 by direct substitution, but factoring the numerator yields (x + 2)(x - 2)/(x - 2). Canceling the common factor leaves lim_{x -> 2} (x + 2) = 4.
Rationalizing by multiplying by a conjugate is effective when square roots are involved. For instance, to evaluate lim_{x -> 0} (sqrt(x + 4) - 2)/x, multiply the numerator and denominator by (sqrt(x + 4) + 2). The numerator simplifies to (x + 4) - 4 = x, which cancels with the x in the denominator, leaving lim_{x -> 0} 1/(sqrt(x + 4) + 2) = 1/4.
Common denominators are useful for complex fractions: combine the fractions in the numerator or denominator, then simplify. Finally, multiplying by a clever form of 1 is especially helpful for trigonometric limits such as lim_{x -> 0} (sin x)/x.
<image>A flowchart for evaluating limits: Start with "Try direct substitution." If the result is a number, done. If the result is 0/0, branch to: "Try factoring," "Try rationalizing," "Try common denominators," or "Try L'Hopital's rule (Lecture 15)." If the result is k/0 (k nonzero), the limit is plus or minus infinity or DNE — check one-sided limits. Each branch leads to "Simplify and try direct substitution again."</image>
III. The Squeeze Theorem
The Squeeze Theorem (also called the Sandwich Theorem or Pinching Theorem) provides a powerful technique for evaluating limits of functions that are difficult to handle directly. It states that if g(x) <= f(x) <= h(x) for all x near a (except possibly at a), and if lim_{x -> a} g(x) = lim_{x -> a} h(x) = L, then lim_{x -> a} f(x) = L. The function f is "squeezed" between g and h, both of which converge to the same limit L, forcing f to converge there as well.
The most important application of the Squeeze Theorem is establishing that lim_{x -> 0} (sin x)/x = 1. This result is proved by showing, through geometric arguments involving the unit circle, that cos x <= (sin x)/x <= 1 for x near 0. Since both cos x and 1 approach 1 as x approaches 0, the Squeeze Theorem forces (sin x)/x to approach 1 as well.
Another instructive application is lim_{x -> 0} x^2 sin(1/x) = 0. Since -1 <= sin(1/x) <= 1 for all nonzero x, it follows that -x^2 <= x^2 sin(1/x) <= x^2. Both bounding functions approach 0, so the squeeze forces the limit to be 0.
<image>A graph illustrating the Squeeze Theorem applied to f(x) = x^2 sin(1/x). The oscillating curve y = x^2 sin(1/x) is plotted in blue, sandwiched between the parabolas y = x^2 (dashed red, upper bound) and y = -x^2 (dashed red, lower bound). Near x = 0, all three curves converge to 0. The region between the bounding parabolas is lightly shaded. Title: "Squeeze Theorem: x^2 sin(1/x) is trapped between -x^2 and x^2."</image>
IV. Continuity at a Point
A function f is continuous at x = a if three conditions are met: first, f(a) must be defined (a is in the domain of f); second, lim_{x -> a} f(x) must exist; and third, lim_{x -> a} f(x) must equal f(a). In plain language, continuity means the limit equals the function value, so there are no gaps, jumps, or holes at the point in question. If any one of these three conditions fails, f is discontinuous at a.
V. Types of Discontinuities
Discontinuities come in several flavors. A removable discontinuity (or hole) occurs when the limit exists but either f(a) is undefined or f(a) does not equal the limit. It can be "fixed" by redefining f(a) to equal the limit. For example, f(x) = (x^2 - 1)/(x - 1) has a removable discontinuity at x = 1.
A jump discontinuity occurs when both one-sided limits exist but are not equal. Piecewise functions and the Heaviside function provide typical examples. An infinite discontinuity occurs when at least one of the one-sided limits is plus or minus infinity, as with f(x) = 1/x at x = 0. An oscillatory discontinuity arises when the function oscillates so wildly that no limit exists, as with sin(1/x) at x = 0.
VI. Continuity on an Interval and Properties
A function f is continuous on an open interval (a, b) if it is continuous at every point in that interval. It is continuous on a closed interval [a, b] if it is continuous on (a, b), continuous from the right at a (meaning lim_{x -> a^+} f(x) = f(a)), and continuous from the left at b (meaning lim_{x -> b^-} f(x) = f(b)).
Several important properties follow from the definition. Combinations of continuous functions are continuous: if f and g are continuous at a, then so are f + g, f - g, f g, c f, and f/g (provided g(a) is not zero). Compositions also preserve continuity: if g is continuous at a and f is continuous at g(a), then f o g is continuous at a. As a consequence, all elementary functions -- polynomials, rational functions, trigonometric, exponential, logarithmic, and root functions -- are continuous on their natural domains.
VII. The Intermediate Value Theorem (IVT)
The Intermediate Value Theorem states that if f is continuous on the closed interval [a, b] and N is any number between f(a) and f(b), then there exists at least one c in (a, b) such that f(c) = N. The intuitive meaning is that a continuous function cannot skip over values: if it goes from f(a) to f(b), it must pass through every value in between.
The most common application of the IVT is proving the existence of roots. If f is continuous on [a, b] and f(a) and f(b) have opposite signs, then there must be some c in (a, b) where f(c) = 0. For example, to show that x^3 + x - 1 = 0 has a root between 0 and 1, observe that f(0) = -1 < 0 and f(1) = 1 > 0. By the IVT, there is a root in (0, 1). It is important to note that the IVT guarantees the existence of c but does not tell you its exact value.
<image>A graph illustrating the Intermediate Value Theorem: a continuous curve y = f(x) on the interval [a, b] with f(a) below the x-axis and f(b) above the x-axis. A horizontal dashed line at y = N (where N is between f(a) and f(b)) intersects the curve at a point c. The point (c, N) is marked with a dot, and a vertical dashed line drops from it to the x-axis at x = c. Labels: f(a), f(b), N on the y-axis; a, c, b on the x-axis. Title: "The Intermediate Value Theorem."</image>
