Continuity and the Intermediate Value Theorem✅

Limits and Continuity

Revising the Idea of a Limit

• Consider a function $y = f(x)$ and a value $l$ on the y-axis.
• For any point $y<em>0$ very close to $l$, if we can find a point $x</em>0$ very close to $a$ (on the x-axis) such that $f(x<em>0) = y</em>0$, then the limit of the function as $x$ approaches $a$ is $l$.
• Formally: If for every $y<em>0$ very close to $l$, we can find an $x</em>0$ very close to $a$, then $\lim_{x \to a} f(x) = l$.

Observation About Limits

• Consider a function $f(x)$ and a value $l$ on the y-axis.
• The limit when $x$ approaches $a$ of $f(x)$ can exist (and be equal to $l$) even if $f(x)$ is not defined at $x = a$, or if $f(a)$ is different from $l$.
• It is not necessarily true that for a limit to exist, it must be equal to the function's value at the point $a$.
• This concept leads to the discussion of continuous functions.

Continuous Functions - Informal Idea

• Informally, a continuous function is one that doesn't have any holes or jumps.
• If there's a hole or a jump in the graph at $x = a$, the function is not continuous at that point.

Revising Specific Cases of Limits

• $\lim_{x \to a} c = c$ (where $c$ is a constant)
• $\lim_{x \to a} x = a$
• $\lim_{x \to a} x^n = a^n$ (where $n$ is a positive integer)
• If $P(x)$ is a polynomial function, then $\lim_{x \to a} P(x) = P(a)$.
• If $R(x)$ is a rational function, and $\lim<em>{x \to a} denominator(R(x)) \neq 0$, then $\lim</em>{x \to a} R(x) = R(a)$.
• $\lim_{x \to a} \sqrt[n]{x} = \sqrt[n]{a}$ (where $n$ is a positive integer, and a > 0 if $n$ is even)

Continuous Functions - Connecting Limits and Function Evaluation

• For continuous functions, the limit of the function as $x$ approaches $a$ can be found by simply substituting $a$ into the function.
• This property makes it easy to determine the limit of a function at a given point.

Examples

Example 1: Direct Substitution (Continuous Function)

• Find $\lim_{x \to 2} (x^2 + 2x - 1)$.
• $\lim<em>{x \to 2} (x^2 + 2x - 1) = \lim</em>{x \to 2} x^2 + \lim<em>{x \to 2} 2x - \lim</em>{x \to 2} 1 = 2^2 + 2(2) - 1 = 4 + 4 - 1 = 7$
• Direct substitution works because the function is a polynomial, and polynomials are continuous.

Example 2: Indeterminate Form (Not Continuous)

• Find $\lim_{x \to 2} \frac{x^2 - 4}{x - 2}$.
• Direct substitution leads to $\frac{0}{0}$, which is not well-defined.
• Factorize: $\frac{x^2 - 4}{x - 2} = \frac{(x + 2)(x - 2)}{x - 2} = x + 2$
• $\lim_{x \to 2} (x + 2) = 2 + 2 = 4$
• The function $y = \frac{x^2 - 4}{x - 2}$ corresponds to a line with a hole at $x = 2$. The limit exists and is equal to $4$, but the function is undefined at $x = 2$. This function is not continuous.

Continuity at a Point

• A function $f$ is continuous at a point $a$ if $\lim_{x \to a} f(x) = f(a)$.
• Conditions for continuity at a point $a$:
  • $f(a)$ must be defined (i.e., $a$ must be in the domain of $f$).
  • $\lim_{x \to a} f(x)$ must exist (the left-hand limit and right-hand limit must be equal).
  • $\lim_{x \to a} f(x) = f(a)$

Examples of Discontinuity

• Condition 1 fails: $f(x) = \frac{1}{x}$ is not continuous at $x = 0$ because $f(0)$ is not defined.
• Condition 2 fails:
  f(x) = \begin{cases}
  -1, & x \leq 0 \
  1, & x > 0
  \end{cases}
  is not continuous at $x = 0$ because $\lim<em>{x \to 0^-} f(x) = -1$ and $\lim</em>{x \to 0^+} f(x) = 1$, so the limit does not exist.
• Condition 3 fails: Let $\lim<em>{x \to 2} f(x) = \lim</em>{x \to 2} \frac{x^2 - 4}{x - 2} = 4$. If we define $f(2) = 2$, then $\lim_{x \to 2} f(x) \neq f(2)$, so $f$ is not continuous at $x = 2$.

Example of Continuity

• $f(x) = x$ is continuous at $x = 2$ because $\lim_{x \to 2} x = 2 = f(2)$.

Continuity in a Domain

• A function $f$ is continuous on an open domain if $f$ is continuous at every point in that domain (e.g., an open interval $(A, B)$).

Properties of Continuous Functions

• Every polynomial is continuous over all real numbers.
• Any rational function is continuous over its domain.
• $f(x) = \sqrt[n]{x}$ is continuous over $[0, \infty)$ if $n$ is a positive integer and continuous over all real numbers if $n$ is an odd positive integer.
• Exponential, sine, and cosine functions are continuous over all real numbers.
• The logarithmic function is continuous over all positive real numbers.

Combination of Functions

• If $f$ and $g$ are continuous at $a$, then:
  • $f + g$ is continuous at $a$.
  • $cf$ is continuous at $a$ (where $c$ is a constant).
  • $fg$ is continuous at $a$.
  • $\frac{f}{g}$ is continuous at $a$ provided that $g(a) \neq 0$.
• If $g$ is continuous at $a$ and $f$ is continuous at $g(a)$, then $f(g(x))$ is continuous at $a$.

Examples

• If $f(x) = e^x$ and $g(x) = \ln x$, then for $x \in (0, \infty)$, $h(x) = e^x + \ln x$ is continuous.
• If $f(x) = \sin x$ and $g(x) = e^x$, then $g(f(x)) = e^{\sin x}$ is continuous at $x = \frac{\pi}{2}$.

Piecewise Functions

• Consider V(t) = \begin{cases}
  \frac{t^2 + 3t - 10}{t - 2}, & t \neq 2 \
  7, & t = 2
  \end{cases}
• For $t \neq 2$, $V(t) = \frac{t^2 + 3t - 10}{t - 2}$ is a rational function and continuous in its largest domain of all real numbers except $2$.
• To check if $V(t)$ is continuous overall, we need to determine if its continuous at $t=2$.
• Then $\lim<em>{t \to 2} V(t) = \lim</em>{t \to 2} \frac{t^2 + 3t - 10}{t - 2} = \lim<em>{t \to 2} \frac{(t + 5)(t - 2)}{t - 2} = \lim</em>{t \to 2} (t + 5) = 2 + 5 = 7 = V(2)$
• Since $\lim_{t \to 2} V(t) = V(2)$, $V(t)$ is continuous at $t = 2$ and therefore continuous for all real numbers.

More on Continuity

• Despite these conditions, there are simple functions that are not continuous at all points.
• Conversely, there are examples of continuous functions that may not seem to be continuous at first sight but are continuous given the above definition.
• If $f(x) = \frac{1}{x}$, the maximal domain is all real numbers except 0, and $f$ is continuous at every point in its maximal domain.
• Consider
  g(x) = \begin{cases}
  x \ln |x|, & x \neq 0 \
  0, & x = 0
  \end{cases}
• Which can be expressed as:
  $g(x) = \begin{cases}<br /> x \ln -x, &amp; x < 0 \ 0, & x = 0 \ x \ln x, & x > 0<br /> \end{cases}$
• From the graph, we can observe that $\lim_{x \to 0} g(x) = 0 = g(0)$. However, to find the limit analytically, we need more advanced calculus principles like L'Hopital's Rule (which will be seen in calculus II).
• From the graph, we can observe that $g(x)$ is continuous for every point on its domain (all real numbers).

Intermediate Value Theorem (IVT)

• If $f$ is a continuous function on a closed interval $[a, b]$ and $f<em>0$ is any value between $f(a)$ and $f(b)$, then there exists a value $c$ in the open interval $(a, b)$ such that $f(c) = f</em>0$.
• The main condition for IVT to apply is that the function must be continuous on the closed interval.

Application of IVT

• Show that there is a number $x$ such that $x = x^3 - 1$.
• Define $f(x) = x^3 - 1 - x$ and consider the domain $[1, 2]$.
• Note that $f(1) = 1^3 - 1 - 1 = -1$ and $f(2) = 2^3 - 1 - 2 = 5$.
• Since $f$ is a polynomial function it is continuous.
• By the IVT, since there exists a value y=0 between $f(1)$ and $f(2)$, there exists an $x<em>0$ between $1$ and $2$ such that $f(x</em>0) = 0$.
• Since substituting $x<em>0$ for $f(x), $x</em>0^3 - 1 - x<em>0 = 0$, that means $x</em>0 = x_0^3 - 1$ .
• We have shown that $x<em>0$ satisfies the conditions that $x</em>0 = x_0^3 - 1$ , with additional proof show that the point exists between 1 and 2.
• The domain can be further reduced. For instance, you can further proceed the algorithm to take the interval, say, [1 ; 1.5] and reapply the method.

Determining Continuity From a Graph

Final examples

• Based on a function's graph:
  • If the function is not defined at that point on its domain, then it is not continuous at that point (e.g., a hole in a graph).
  • If the limit does not exist at the point, then the function is not continuous at that point (e.g., different limits when approaching from the left and right, or an asymptote in the graph.
  • Even if the graph looks continuous, the function may not be continuous if there is not clear values to compare the limit to. More tests may need to take place.
  • If the limit when x approaches two of the function in x is the value of the function evaluated at the same point, then the function is continuous.
• Then list all of the functions on the domain R that are continuous at said points, and which are only continuous on a small subset of it.