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Calculus 1

Continuity at a point

A limit predicts the height a function heads toward near an input, and the value says where the function actually is at that input. When do the prediction and the value agree, and how can you tell without drawing anything?

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ContinuityGrade 11The limit lawsCalculus 1
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Continuity on an intervalCalculus 1Making a piecewise function continuousCalculus 1
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Continuity at a point
Continuity on an intervalDiscontinuities and how to remove themThe extreme value theoremMaking a piecewise function continuous

Why this exists

A limit and a value answer two different questions. The limit \lim_{x \to a} f(x) looks only at inputs near a, never at a itself, and says what height the outputs are heading toward. The value f(a) is what the function actually does at a. In The limit laws the two kept agreeing for polynomials and for rational functions with a nonzero bottom, which is why direct substitution worked; and in Limits by algebra: factoring and rationalizing they kept disagreeing, because the function was not even defined at the point.

A function whose value at a is exactly the height its limit predicts is called continuous at a. At such a point the limit is found by plugging in, and continuity is the hypothesis behind the big theorems of calculus. The picture of a graph drawn without lifting the pencil is in Continuous graphs, jumps and holes; this lesson makes that picture exact with limits and works out its algebra.

Definition

In plain words: a function is continuous at a when its limit at a exists and equals its value at a.

In symbols, f is continuous at a when

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

This one equation carries three conditions, and all three must hold:

  • Condition 1: f(a) is defined, so a is in the domain of f.
  • Condition 2: \lim_{x \to a} f(x) exists as a number. By One sided limits and when a limit exists, this means both one-sided limits exist and are equal.
  • Condition 3: the two numbers are equal, \lim_{x \to a} f(x) = f(a).

If f is defined on both sides of a and any of the three fails, f is discontinuous at a, and a is a discontinuity of f. Here f is assumed to be defined at every input in some open interval around a, except perhaps at a itself, so that the two-sided limit makes sense. At an end of a domain only one side can be watched; that case belongs to Continuity on an interval.

Intuition

Think of the limit as a prediction made by someone who can see every value of f except the one at a. Watching f(x) as x closes in on a from both sides, they predict “the output at a should be L”. Continuity at a says the prediction comes true: the function is there, and it is exactly where its neighbors said it would be.

Each condition is a way the prediction can go wrong. If Condition 1 fails, there is nothing at a to check. If Condition 2 fails, the neighbors disagree or never settle, so there is no single prediction. If Condition 3 fails, the value sits somewhere other than the prediction. Figure 1 shows the third case beside the case where all is well.

2 4 4 5 x y
f(2) = 4
(2, 5)
 A 2 4 4 5 x y
f(2) = 5
 B
Figure 1The function of Example 1 with f(2) = 4 (A) and with f(2) = 5 (B). In both the limit at 2 is 5; only in B does the value match it.

Because the limit ignores the value at a, changing f(a) alone can never repair a failure of Condition 2, but it always repairs a failure of Condition 1 or 3. Which breaks can be repaired this way, and how, is taken up in Discontinuities and how to remove them.

Results

Polynomials and rational functions. Every polynomial is continuous at every number. A rational function \dfrac{p(x)}{q(x)} is continuous at every a with q(a) \ne 0, and discontinuous where q(a) = 0, since Condition 1 fails there.

Other basic functions (stated without proof). \displaystyle \sqrt[n]{x} is continuous at every a in its domain other than an end of it (so at every a > 0 when n is even, and at every a when n is odd), and \sin x, \cos x and b^x (b > 0) are continuous at every number.

Sums, products and quotients. If f and g are both continuous at a and k is a constant, then these are continuous at a:

f + g, \qquad f - g, \qquad k f, \qquad f g

and so is \dfrac{f}{g}, provided g(a) \ne 0.

Compositions. If g is continuous at a and f is continuous at g(a), then f \circ g, the function x \mapsto f\big(g(x)\big), is continuous at a:

\lim_{x \to a} f\big(g(x)\big) = f\big(g(a)\big)
Why it is true

Sums. Suppose f and g are continuous at a, so \lim_{x \to a} f(x) = f(a) and \lim_{x \to a} g(x) = g(a). Both limits exist, so the sum law applies:

\begin{aligned} \lim_{x \to a} \big[f(x) + g(x)\big] &= \lim_{x \to a} f(x) + \lim_{x \to a} g(x) \\ &= f(a) + g(a) \end{aligned}

The last line is the value of f + g at a, which is defined because f(a) and g(a) are. All three conditions hold for f + g. The difference, constant multiple and product results are the same argument with the difference, constant multiple and product laws.

Quotients. With g(a) \ne 0, the bottom limit \lim_{x \to a} g(x) = g(a) is not 0, so the quotient law gives \displaystyle \lim_{x \to a} \frac{f(x)}{g(x)} = \frac{f(a)}{g(a)}, which is the value of \displaystyle \frac{f}{g} at a. The quotient law is exactly where g(a) \ne 0 is needed.

Polynomials and rational functions. Constants and x are continuous everywhere, since \lim_{x \to a} c = c and \lim_{x \to a} x = a. A polynomial is built from these by multiplying and adding, and a rational function is a quotient of polynomials, so the product, sum and quotient results finish the proof.

Compositions. Since g is continuous at a, \lim_{x \to a} g(x) = g(a). The limit law for a composition, from The limit laws, says that when the inner limit is L and f is continuous at L, the limit of f\big(g(x)\big) is f(L). With L = g(a) this gives \lim_{x \to a} f\big(g(x)\big) = f\big(g(a)\big), the value of f \circ g at a.

Worked examples

Example 1

Alsu’s two choices for one missing value

Let f(x) = \dfrac{x^2 + x - 6}{x - 2} for x \ne 2. Alsu completes the definition in two ways: first with f(2) = 4, then with f(2) = 5. Is f continuous at 2 in each case?

Solution

The limit does not depend on f(2), so find it once. Substituting x = 2 into the formula gives \displaystyle \frac{0}{0}, so factor: x^2 + x - 6 = (x + 3)(x - 2). For x \ne 2 the factor x - 2 is not zero and cancels, leaving f(x) = x + 3. The limit looks only at x \ne 2, so

\lim_{x \to 2} f(x) = \lim_{x \to 2} (x + 3) = 5

A table agrees: f(1.99) = 4.99 and f(2.01) = 5.01.

With f(2) = 4: Condition 1 holds, since f(2) is defined. Condition 2 holds, the limit is 5. Condition 3 fails, since 5 \ne 4. So f is not continuous at 2; the graph is the line y = x + 3 with the point above 2 moved down to height 4 (Figure 1, A).

With f(2) = 5: all three conditions hold, since \lim_{x \to 2} f(x) = 5 = f(2). So f is continuous at 2. Now f(x) = x + 3 for every x, and its graph is the whole line (Figure 1, B).

Example 2

Ildar’s value with no limit behind it

Let k(x) = \dfrac{x}{(x - 4)^2} for x \ne 4, and k(4) = 0. Ildar asks whether k is continuous at 4.

Solution

Condition 1 holds, since k(4) = 0. For Condition 2, watch the formula near 4. The top heads to 4, while the bottom heads to 0 and is positive on both sides of 4, because it is a square. A table shows the effect:

\begin{aligned} k(3.99) &= \frac{3.99}{0.0001} = 39{,}900 \\ k(4.01) &= \frac{4.01}{0.0001} = 40{,}100 \end{aligned}

Closer in, \displaystyle k(4.001) = \frac{4.001}{0.000001} = 4{,}001{,}000. The outputs grow without bound from both sides, which is written \lim_{x \to 4} k(x) = \infty. That describes the behavior; it is not a number, so the limit does not exist. Condition 2 fails and k is not continuous at 4. No choice of k(4) could repair this: there is no height for the value to match.

Example 3

Rustem’s rational function, point by point

At which numbers is q(x) = \dfrac{x^2 + 1}{x^2 - 2x - 8} continuous? Find \lim_{x \to 1} q(x).

Solution

The top and bottom are polynomials, so both are continuous at every number. By the quotient result, q is continuous at every a where the bottom is not 0. The bottom factors as x^2 - 2x - 8 = (x - 4)(x + 2), which is 0 at x = 4 and x = -2 only. There q is not defined, so Condition 1 fails; at every other number q is continuous.

Since 1 is neither 4 nor -2, q is continuous at 1, and the limit is the value:

\lim_{x \to 1} q(x) = q(1) = \frac{1 + 1}{1 - 2 - 8} = -\frac{2}{9}
Example 4

Values instead of limits

Functions f and g are continuous at 2, with f(2) = 3 and g(2) = -1. Find

\lim_{x \to 2} \frac{\big(f(x)\big)^2 g(x) - 4 f(x)}{g(x) + 3}
Solution

Only values are given, but continuity turns them into limits: \lim_{x \to 2} f(x) = 3 and \lim_{x \to 2} g(x) = -1. Better still, the results build continuity of the whole expression. The product result makes f^2 g = f \cdot f \cdot g continuous at 2, the constant multiple and difference results make the top continuous at 2, and the sum result makes g + 3 continuous at 2. The bottom’s value is g(2) + 3 = 2 \ne 0, so the quotient result makes the whole fraction continuous at 2, and its limit is its value:

\begin{aligned} \frac{3^2 \cdot (-1) - 4 \cdot 3}{-1 + 3} &= \frac{-9 - 12}{2} \\ &= -\frac{21}{2} \end{aligned}
Example 5

A square root of a quotient

Let \displaystyle F(x) = \sqrt{\dfrac{x + 1}{x - 3}}. Show that F is continuous at 5 and find F(5). Is F continuous at 2?

Solution

F is a composition: the inner function \displaystyle g(x) = \frac{x + 1}{x - 3} feeds the outer function \displaystyle f(u) = \sqrt{u}. The inner function is rational with bottom 5 - 3 = 2 \ne 0 at 5, so g is continuous at 5, with \displaystyle g(5) = \frac{6}{2} = 3. The square root is continuous at 3, since 3 > 0. By the composition result, F is continuous at 5, and

F(5) = \sqrt{3}

At 2 the inner value is \displaystyle g(2) = \frac{3}{-1} = -3, and \displaystyle \sqrt{-3} is not a real number. So F(2) is not defined and Condition 1 fails: F is not continuous at 2. The composition result does not apply, because the outer function is not continuous at -3; it is not even defined there.

Example 6

A continuous product of a broken factor

Let s(x) = 1 for x \ge 0 and s(x) = -1 for x < 0. Is p(x) = x\,s(x) continuous at 0? Is r(x) = (x + 1)\,s(x)?

Solution

The factor s jumps at 0, so the product result says nothing; check the conditions directly.

For p: p(0) = 0 \cdot 1 = 0, so Condition 1 holds. For x > 0, p(x) = x, and for x < 0, p(x) = -x, so both one-sided limits are 0 and \lim_{x \to 0} p(x) = 0. That equals p(0), so p is continuous at 0. In fact p(x) = |x|: the factor x shrinks to 0 and squeezes out the jump in s.

For r: r(0) = 1. For x > 0, r(x) = x + 1 \to 1; for x < 0, r(x) = -(x + 1) \to -1. The one-sided limits differ, so Condition 2 fails and r is not continuous at 0.

So a product with a broken factor may be continuous (p) or not (r): the product result is silent when its hypothesis fails.

Boundaries

  • The definition needs inputs on both sides. The two-sided limit, and so this definition, applies at an a with the function defined on both sides of it. At an end of a domain, as \displaystyle \sqrt{x} at 0, continuity is defined with a one-sided limit in Continuity on an interval.
  • The quotient result needs g(a) \ne 0. Where the bottom is 0, the quotient is not defined and cannot be continuous. Example 3 shows this at two inputs.
  • The composition result needs the outer function continuous at the inner value g(a), not at a. Example 5 and Exercise 16 show it failing.
  • The results go one way only. They give continuity when their hypotheses hold. When a hypothesis fails, the combination still has to be checked, and it may turn out continuous, as p in Example 6 does.
  • Continuity at a point says nothing about nearby points or about smoothness. A function can be continuous at a and nowhere else near it. And |x| is continuous at 0 even with its sharp corner.

Where this goes next

Continuity at every point of an interval, with one-sided continuity at its ends, is Continuity on an interval; it is the hypothesis of the The extreme value theorem and of the intermediate value theorem, which says that an unbroken graph takes every height in between. Sorting discontinuities into removable ones and jumps, and repairing the removable ones, is Discontinuities and how to remove them. Choosing a constant so that the pieces of a piecewise function meet is Making a piecewise function continuous. Further on, the derivative at a point is a limit, and a function with a derivative at a turns out to be continuous at a (see The derivative at a point).

Exercises

The three conditions

  1. Let f(x) = \dfrac{x^2 + 4x - 21}{x - 3} for x \ne 3, and f(3) = 10. Is f continuous at 3?
  2. Let g(x) = \dfrac{2x^2 + x - 1}{x + 1} for x \ne -1, and g(-1) = 3. Is g continuous at -1? If not, which condition fails?
  3. Let h(x) = \dfrac{x^2 - 5x}{x - 5}. Find \lim_{x \to 5} h(x), and decide whether h is continuous at 5.
  4. Let k(x) = x^2 - 3 for x < 2 and k(x) = 2x - 3 for x \ge 2. Is k continuous at 2?
  5. Let m(x) = 4 - x for x < 1 and m(x) = x^2 + 1 for x \ge 1. Is m continuous at 1? If not, which condition fails?
  6. For the function w graphed below, decide at each of a = -3, -1, 1 and 3 whether w is continuous. Where it is not, name the first of the three conditions that fails.
−5 −3 −1 1 3 5 −2 3 5 x y
Figure 2The graph of w for Exercise 6.

Building continuous functions

  1. At which numbers is \dfrac{x^2 + 4}{x^2 - 7x + 10} continuous?
  2. Explain why \dfrac{3x - 1}{x^2 + 4} is continuous at every number.
  3. Functions f and g are continuous at 1, with f(1) = 4 and g(1) = -2. Find \lim_{x \to 1} \dfrac{f(x)\,g(x) + 3\big(g(x)\big)^2}{f(x) - g(x)}.
  4. With f and g as in Exercise 9, does the quotient result show that \dfrac{f}{g + 2} is continuous at 1? Is it continuous at 1?
  5. Use continuity to find \lim_{x \to 3} \dfrac{x^3 - 2x}{x + 1}.

Compositions

  1. Show that \displaystyle \sqrt{x^2 - 3x + 5} is continuous at 4, and find its value there.
  2. Find \lim_{x \to 2} (x^2 - 1)^5, naming the result that allows substitution.
  3. Let \displaystyle K(x) = \dfrac{1}{\sqrt{2x + 1}}. Is K continuous at 4? At -1?
  4. Find \lim_{x \to 1} \cos\!\left(\dfrac{\pi x^2}{3}\right), using that \cos is continuous everywhere.
  5. Let f(u) = 1 for u < 0 and f(u) = 3 for u \ge 0, and let g(x) = x^2 - 4. Is f\big(g(x)\big) continuous at 2? At 0?

Reasoning

  1. Give functions f and g, each discontinuous at 0, whose sum f + g is continuous at 0.
  2. Alsu says: “If f is continuous at a and g is not, then f + g is not continuous at a.” Is she right? Explain.
  3. Ildar says: “If f\big(g(x)\big) is continuous at a, then g is continuous at a.” Is he right? Explain.
  4. Explain why the single equation \lim_{x \to a} f(x) = f(a) already contains all three conditions.
  5. Prove that if f and g are continuous at a, then f g is continuous at a.
  6. Rustem says f is continuous at a exactly when \lim_{h \to 0} f(a + h) = f(a). Explain why he is right.
Answers
  1. Yes. For x \ne 3, \displaystyle \frac{(x + 7)(x - 3)}{x - 3} = x + 7, so the limit is 3 + 7 = 10 = f(3). All three conditions hold.
  2. No; Condition 3 fails. For x \ne -1, \displaystyle \frac{(2x - 1)(x + 1)}{x + 1} = 2x - 1, so the limit is -3, while g(-1) = 3.
  3. The limit is 5; h is not continuous at 5. For x \ne 5, \displaystyle \frac{x(x - 5)}{x - 5} = x, so the limit is 5, but h(5) is not defined (\displaystyle \frac{0}{0}), so Condition 1 fails.
  4. Yes. The left limit is 2^2 - 3 = 1, the right limit is 2 \cdot 2 - 3 = 1, and k(2) = 1.
  5. No; Condition 2 fails. m(1) = 2 is defined, but the left limit is 4 - 1 = 3 and the right limit is 1 + 1 = 2, so the limit does not exist.
  6. At -3: Condition 1 fails; there is an open circle and no dot, so w(-3) is not defined (the limit is 2). At -1: Condition 2 fails; w(-1) = 3, but the left limit is 3 and the right limit is 1. At 1: Condition 3 fails; the limit is 3 but w(1) = -2. At 3: continuous; w(3) = 5 and both sides head to 5, the corner does not matter.
  7. At every number except 2 and 5. The bottom is (x - 2)(x - 5), zero only at 2 and 5, where the function is undefined; elsewhere the quotient result applies.
  8. Top and bottom are polynomials, continuous everywhere, and the bottom is x^2 + 4 \ge 4, never 0, so the quotient result applies at every number.
  9. \displaystyle \frac{2}{3}. The expression is continuous at 1 (bottom value 4 - (-2) = 6 \ne 0), so its limit is its value: \displaystyle \frac{4(-2) + 3 \cdot 4}{6} = \frac{4}{6} = \frac{2}{3}.
  10. No, and no. The bottom has value g(1) + 2 = 0, so the quotient result does not apply; and \displaystyle \frac{f}{g + 2} is not defined at 1, so Condition 1 fails.
  11. \displaystyle \frac{21}{4}. The bottom is 4 \ne 0 at 3, so the function is continuous there: \displaystyle \frac{27 - 6}{4} = \frac{21}{4}.
  12. Value 3. The inner polynomial is continuous at 4 with value 16 - 12 + 5 = 9, and \displaystyle \sqrt{u} is continuous at 9 > 0; by the composition result the function is continuous at 4, with value \displaystyle \sqrt{9} = 3.
  13. 243. u^5 is continuous at u = 3 and x^2 - 1 is continuous at 2 with value 3, so the composition result gives 3^5 = 243.
  14. Yes at 4, no at -1. At 4 the inner value is 9, \displaystyle \sqrt{9} = 3 \ne 0, so \displaystyle K(4) = \frac{1}{3} and the root and quotient results apply. At -1 the inner value is -1, so K(-1) is not defined and Condition 1 fails.
  15. \displaystyle \frac{1}{2}. The inner function \displaystyle \frac{\pi x^2}{3} is continuous with value \displaystyle \frac{\pi}{3} at 1, and \displaystyle \cos\frac{\pi}{3} = \frac{1}{2}.
  16. Not at 2; yes at 0. Near 2 with x < 2, g(x) < 0 and the composition is 1; with x > 2, g(x) > 0 and it is 3; the one-sided limits differ, so Condition 2 fails. At 0, g(0) = -4 and f is constant near -4, so it is continuous there; the composition is continuous at 0 with value 1.
  17. One example: f = s from Example 6 and g = -s. Each jumps at 0, but f + g = 0 everywhere, which is continuous.
  18. Yes. If f + g were continuous at a, then g = (f + g) - f would be a difference of two functions continuous at a, hence continuous at a by the difference result, which it is not.
  19. No. Take f(u) = 7 for every u and g = s from Example 6. Then f\big(g(x)\big) = 7 is continuous at 0, but g is not.
  20. For the equation to make sense its right side f(a) must exist (Condition 1) and its left side must be a number (Condition 2); the equation then says they are equal (Condition 3).
  21. \lim_{x \to a} f(x) = f(a) and \lim_{x \to a} g(x) = g(a) both exist, so the product law gives \lim_{x \to a} f(x)\,g(x) = f(a)\,g(a), which is the value of f g at a.
  22. Put x = a + h. Then x \to a exactly when h \to 0, with x \ne a exactly when h \ne 0, so \lim_{x \to a} f(x) and \lim_{h \to 0} f(a + h) are the same limit, and each equation says it equals f(a).
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