Continuity at a Point | Math Coach Amy

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Key Ideas Easy Medium Hard

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

value exists, limit exists, they match

What to Remember

Three things

f is continuous at a when

1. \(f(a)\) exists

2. \(\lim_{x \to a} f(x)\) exists

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

One-sided at an endpoint

On a closed interval, continuity at the left endpoint only needs the right-hand limit. At the right endpoint, only the left-hand limit.

Continuous on an interval

Every point of the interval must be a continuity point. A single hole, jump, or vertical asymptote on that interval is enough to say no.

Polynomials and rationals

A polynomial is continuous everywhere. A rational is continuous on its domain — wherever the denominator is not 0.

Level 1

Easy

Check \(f(a)\), the limit, and whether they match.

Example

Let \(f(x) = \begin{cases} x + 3 & -2 \leq x < 0 \\ 3 - 2x & 0 < x \leq 2 \\ 1 & x = 0 \end{cases}\). Does \(f(0)\) exist?

The middle listing says \(f(0) = 1\).

Yes.

Answer: yes

Let \(f(x) = \begin{cases} x + 3 & -2 \leq x < 0 \\ 3 - 2x & 0 < x \leq 2 \\ 1 & x = 0 \end{cases}\). Does \(\lim_{x \to 0} f(x)\) exist?

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Left: \(0 + 3 = 3\).

Right: \(3 - 0 = 3\).

Yes. The limit is 3.

Answer: yes

Let \(f(x) = \begin{cases} x + 3 & -2 \leq x < 0 \\ 3 - 2x & 0 < x \leq 2 \\ 1 & x = 0 \end{cases}\). Is \(\lim_{x \to 0} f(x) = f(0)\)?

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\(\lim_{x \to 0} f(x) = 3\)

\(f(0) = 1\)

No.

Answer: no

Let \(f(x) = \begin{cases} x + 3 & -2 \leq x < 0 \\ 3 - 2x & 0 < x \leq 2 \\ 1 & x = 0 \end{cases}\). Is \(f\) continuous at \(x = 0\)?

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The limit exists, and \(f(0)\) exists, but they are not equal.

No.

Answer: no

Let \(g(x) = \begin{cases} 2x + 1 & x < 3 \\ 7 & x = 3 \\ x + 4 & x > 3 \end{cases}\). Is \(g\) continuous at \(x = 3\)?

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Left: \(6 + 1 = 7\).

Right: \(3 + 4 = 7\).

\(g(3) = 7\)

Yes.

Answer: yes

Is \(h(x) = x^2 - 4x\) continuous at \(x = 2\)?

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Polynomials are continuous everywhere.

Yes.

Answer: yes

Level 2

Medium

A piecewise with several candidates, or continuity on an interval.

Example

Let \(f(x) = \begin{cases} x + 5 & x < 1 \\ 6 & x = 1 \\ 8 - 2x & 1 < x < 4 \\ 1 & x = 4 \\ x - 4 & x > 4 \end{cases}\). Which statement is false?

\(f\) is continuous at \(x = 1\).

\(f\) is continuous at \(x = 2\).

\(f\) is continuous at \(x = 4\).

\(f\) is continuous at \(x = 5\).

At 1: left \(6\), right \(6\), \(f(1) = 6\). Continuous.

At 2: only the middle piece, \(8 - 4 = 4\). Continuous.

At 4: left \(8 - 8 = 0\), \(f(4) = 1\). The limit is 0 if the right side is also 0: right \(4 - 4 = 0\). Limit is 0, not equal to 1. Not continuous.

At 5: \(5 - 4 = 1\). Continuous.

The false statement is “\(f\) is continuous at \(x = 4\).”

Answer: \(f\) is not continuous at \(x = 4\)

Which functions are continuous on \(0 < x < 5\)?

I. \(f(x) = \dfrac{x - 4}{x - 7}\)

II. \(g(x) = \dfrac{1}{x^2 - 9}\)

III. \(h(x) = \ln(x + 2)\)

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I. Denominator 0 at 7, which is not in \((0, 5)\). Continuous on the interval.

II. Denominator 0 at \(x = 3\), which is in the interval. Not continuous on the interval.

III. \(\ln(x + 2)\) needs \(x > -2\), so it is continuous on \((0, 5)\).

I and III only.

Answer: I and III only

Which function is not continuous on \((-\infty, \infty)\)?

\(p(x) = 4x^2 - 2x + 1\)

\(q(x) = \dfrac{1}{x^3 + x^2 - 6x}\)

\(r(x) = \cos(\pi x)\)

\(s(x) = \dfrac{1}{e^x}\)

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\(4x^2 - 2x + 1\) is a polynomial.

\(\cos(\pi x)\) is continuous everywhere.

\(\dfrac{1}{e^x}\) is never 0 in the denominator.

\(\dfrac{1}{x^3 + x^2 - 6x}\) is a rational whose denominator is a cubic, so it has at least one real zero.

The rational function is not continuous on all real numbers.

Answer: \(\dfrac{1}{x^3 + x^2 - 6x}\)

Let \(f(x) = \dfrac{x^2 - 1}{x - 1}\) for \(x \neq 1\). Why is \(f\) not continuous at \(x = 1\)?

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The denominator is 0 at \(x = 1\), so \(f(1)\) does not exist.

After canceling, \(\lim_{x \to 1} f(x) = 2\) does exist.

The limit exists, but \(f(1)\) does not.

Answer: the limit exists, but \(f(1)\) does not

Let \(f(x) = \begin{cases} x^2 & x \leq 2 \\ 6 - x & x > 2 \end{cases}\). Is \(f\) continuous at \(x = 2\)?

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Left: \(4\).

Right: \(4\).

\(f(2) = 4\)

Yes.

Answer: yes

Is \(f(x) = \dfrac{x - 3}{x^2 - 9}\) continuous at \(x = 3\)?

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\(f(3)\) is undefined.

No.

Answer: no

Level 3

Hard

Justify with limit notation at more than one x-value.

Example

Let \(k(x) = \bigl(5 - f(x)\bigr)g(x)\). The graph of \(f\) is shown, and \(g\) is continuous with \(g(-1) = 2\) and \(g(1) = 3\). Is \(k\) continuous at \(x = 1\)?

Graph of f

\(f\) is continuous at 1 with \(f(1) = 2\).

\(g\) is continuous at 1.

A product and difference of continuous functions is continuous, so \(k\) is continuous at 1.

\(k(1) = (5 - 2)(3) = 9\)

Answer: yes

Let \(k(x) = \bigl(5 - f(x)\bigr)g(x)\). The graph of \(f\) is shown, and \(g\) is continuous. Is \(k\) continuous at \(x = -1\)?

Graph of f

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\(\lim_{x \to -1} f(x)\) does not exist (left 4, right 2).

Then \(\lim_{x \to -1} k(x)\) does not exist.

No.

Answer: no

Let \(f(x) = \dfrac{x^3 + 2x - 12}{4\cos\bigl(\frac{\pi}{2}x\bigr) + x^2}\). Why is \(f\) not continuous at \(x = 2\)?

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Denominator at 2: \(4\cos(\pi) + 4 = 4(-1) + 4 = 0\).

So \(f(2)\) does not exist.

The two-sided limit can still exist (a hole).

The limit exists, but \(f(2)\) does not.

Answer: the limit exists, but \(f(2)\) does not

On \([-2, 3]\), a function is continuous at every x except a hole at 0. Is it continuous on \([-2, 3]\)?

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Continuous on a closed interval means continuous at every point of the interval, including one-sided continuity at the endpoints.

A hole at 0 is inside the interval.

No.

Answer: no

Let \(f\) be increasing for \(x < 2\) and increasing for \(x > 2\). If \(\lim_{x \to 2} f(x) = 4\), which could represent \(f\)?

I. \(f(x) = \dfrac{x^2 - 4}{x - 2}\) for \(x \neq 2\), and \(f(2) = 1\)

II. A table whose y-values blow up to \(\pm \infty\) as \(x\) approaches 2

III. A graph with a hole at \((2, 4)\) and a filled point at \((2, 1)\)

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I. \(\dfrac{(x - 2)(x + 2)}{x - 2} = x + 2 \to 4\). Matches the limit. The function is increasing on each side of 2.

II. Blowing up means the limit does not exist.

III. The graph approaches 4, so the limit is 4.

I and III only.

Answer: I and III only

A table of \(f\) near \(x = -2\) has y-values approaching 3 from both sides, and \(f(-2) = 1\). Which table feature is required?

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The nearby y-values must approach 3, not 1.

The value at \(x = -2\) should be 1, not 3, and should not be blank if the function is defined there.

Nearby values near 3, with \(f(-2) = 1\).

Answer: nearby values approach 3, and \(f(-2) = 1\)

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