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PAGE 1

MA 161 Exam 1

Tuesday, 2/1/2022, 6:30 pm

Recitation InstructorLocation
Isaac ChiuCL50 224
Rowan DesjardinsCL50 224
Christopher DeweyCL50 224
Hanan GadiWALC 1055
Mohit PathakWALC 1055
Vittal SrinivasanPHYS 112
Ezekiel (Seun) Yinka KehindePHYS 112
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2.6 Continuity

If a function is continuous, then there are no breaks in its graph.

  • If tracing you never need to lift hand to complete the trace.
Coordinate graph showing a continuous parabola y = x^2 .

\( y = x^2 \)

Coordinate graph showing a continuous sine wave y = \sin x .

\( y = \sin x \)

Mathematically, if a function is continuous at \( x = a \) then:

  1. \( f(a) \) is defined
  2. \( \lim_{x \to a} f(x) \) exists
  3. \( f(a) = \lim_{x \to a} f(x) \)

ALL 3 need to happen

for \( f(x) \) to be continuous at \( x = a \)

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Infinite Discontinuity

\[ f(x) = \frac{-1}{1-x} \]

\( f(1) \) is not defined because of the asymptote and \( \lim_{x \to 1} f(x) \) DNE (Does Not Exist).

So \( f(x) = \frac{-1}{1-x} \) is discontinuous at \( x=1 \) because the first two requirements are not met, but is continuous at all \( x \neq 1 \).

Graph of f(x) = -1/(1-x) showing a vertical asymptote at x=1 and an infinite discontinuity.

infinite discontinuity

Removable Discontinuity

\[ f(x) = \frac{x^2+x-20}{x-4} = \frac{(x+5)(x-4)}{x-4} = x+5, \quad x \neq 4 \]

\( f(4) \) is not defined. So \( f(x) \) is discontinuous at \( x=4 \).

Linear graph with a hole at x=4, labeled as a removable discontinuity.
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The hole is called a removable discontinuity because we can define \( f(x) \) there to "patch" the hole.

\[ f(x) = \begin{cases} \frac{x^2+x-20}{x-4} & \text{if } x \neq 4 \\ 9 & \text{if } x = 4 \end{cases} \]

The value 9 is chosen to fill the hole in the graph.

Graph of the piecewise function where the hole at (4,9) is filled with a solid point.

Jump Discontinuity

Another type of discontinuity: jump discontinuity

Graph showing a jump discontinuity at x=1 where the function values abruptly change.
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One-Sided Continuities

Just like with limits, we can talk about one-sided continuities.

\[ f(x) = \begin{cases} x + 3 & x \le 0 \\ x^2 + 5 & x > 0 \end{cases} \]

discontinuous at \( x = 0 \)

but is continuous from the LEFT at \( x = 0 \)

because \[ \lim_{x \to 0^-} f(x) = f(0) \]

but not continuous from the RIGHT at \( x = 0 \)

because \[ \lim_{x \to 0^+} f(x) \neq f(0) \]

\( 5 \)

\( 3 \)

Graph of a piecewise function with a jump discontinuity at x=0. A line x+3 ends at (0,3) and a parabola x^2+5 starts at (0,5).
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Continuity of Common Functions

many common functions are continuous on their domains

  • polynomials
  • rationals
  • root
  • trig
  • exponential
  • log

wherever defined, \[ f(a) = \lim_{x \to a} f(x) \]

\[ f(a) = \lim_{x \to a} f(x) \] due to continuity is the reason why we can plug in \( x = a \) into \( f(x) \) to find \( \lim_{x \to a} f(x) \)

\[ \lim_{x \to 1} (x^2 + 2x - 1) = (1)^2 + (2)(1) - 1 = 2 \]

polynomial

continuous everywhere

so \( \lim_{x \to 1} f(x) = f(1) \)

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Limits of Continuous Functions

\[ \lim_{x \to 3} \ln(5x) = \ln(5 \cdot 3) = \ln 15 \]

Analysis:

  • Log function defined for \( x > 0 \)
  • So continuous for \( x > 0 \)
  • So, \( \lim_{x \to 3} f(x) = f(3) \)
\[ \lim_{x \to 0} (5e^x + \cos(\pi x) + \sqrt{5x^2 + 10}) = 5e^0 + \cos(\pi \cdot 0) + \sqrt{5 \cdot 0 + 10} \]\[ = 6 + \sqrt{10} \]

Exponential

Continuous on \( (-\infty, \infty) \)

Trig

Continuous on \( (-\infty, \infty) \)

Root function

Continuous on \( 5x^2 + 10 \ge 0 \)

\( x \to 0 \): 0 is in interval of continuity for all parts.

So, \( \lim_{x \to 0} f(x) = f(0) \)

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Example: Continuity of Piecewise Functions

\[ f(x) = \begin{cases} cx^2 + 2x & \text{if } x < 2 \\ x^3 - cx & \text{if } x \ge 2 \end{cases} \]

Find \( c \) so \( f(x) \) is continuous everywhere.

Both pieces are polynomial so continuous on their own domains.

So \( f(x) \) is continuous everywhere except possibly at \( x = 2 \).

For \( f(x) \) to be continuous at \( x = 2 \)

Need:

  1. \( f(2) \) defined Yes, \( f(2) = 8 - 2c \)
  2. \( \lim_{x \to 2} f(x) \) exist \( \to \lim_{x \to 2^-} f(x) = \lim_{x \to 2^+} f(x) \)
  3. \( \lim_{x \to 2} f(x) = f(2) \)
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Continuity and Limits Analysis

\[ \lim_{x \to 2^+} f(x) = \lim_{x \to 2^+} x^3 - cx = 8 - 2c \]

Note: This limit applies for \( x > 2 \).

\[ \lim_{x \to 2^-} f(x) = \lim_{x \to 2^-} cx^2 + 2x = 4c + 4 \]

Note: This limit applies for \( x < 2 \).

For these to be equal:

\[ 8 - 2c = 4c + 4 \implies 6c = 4 \implies c = 2/3 \]

If \( c = 2/3 \), then 2) \( \lim_{x \to 2} f(x) \) exists is true.

Checking Continuity Condition 3

Now check 3) \( \lim_{x \to 2} f(x) = f(2) \)

Use \( c = 2/3 \) in \( 8 - 2c \) or \( 4c + 4 \):

\[ \frac{20}{3} = 8 - 2c \quad (\text{where } c = 2/3) \]
\[ = \frac{20}{3} \]

So, if \( c = 2/3 \), then \( f(x) \) is continuous everywhere.

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Intermediate Value Theorem (IVT)

If \( f(x) \) is continuous on \( [a, b] \), then \( f(x) \) takes on every value between \( f(a) \) and \( f(b) \).

Example

For example, \( f(x) = x^2 \) on \( [0, 2] \).

Graph of f(x)=x^2 from x=0 to x=2. The y-axis goes from 0 to 4. A green curve starts at (0,0) and ends at (2,4).

IVT says the function takes on every value between 0 and 4.

This is because a continuous function has no breaks, so \( f(x) = x^2 \) cannot skip any value between 0 and 4.