02. Continuity
May 1 Lecture. Links to the board and the recording.
Definition:¶
\(f(x)\text{ is continuous at }x=a\text{ if }f(a)=\lim_{x\to a}{f(x)}\).
3 Types of discontinuities:¶
- \(\lim_{x\to a}{f(x)}\) exists, but \(\not=f(a)\). Either \(f(a)\) is undefined, or some other number.
- This is called a removable discontinuity, because it’s an issue at just that point.
- \(\lim_{x\to a}{f(x)}\) doesn’t exist because \(\lim_{x\to a^-}{f(x)} \not=\lim_{x\to a^+}{f(x)}\)
- This is called a jump discontinuity, or a Type I discontinuity.
- \(\lim_{x\to a}{f(x)}\) doesn’t exist, even one-sided (from at least one side).
- This is called a Type II discontinuity.
- This does not include extended limits, so by this definition, \(\lim_{x\to0}{\frac1x}\) doesn’t have a limit, meaning it’s discontinuous at \(x=0\) with a Type II discontinuity.
- Usually in this case, \(f\) has a vertical asymptote \(x=a\), but not always.
Examples of discontinuities
- \(f(x) = {1\over x+3} \text{ at } x=-3\)
- \(f(x)=\ln(x) \text{ at }x=0\)
- \(f(x)=\sin{\frac 1x}\)
- Type II @ x=0 because the limit doesn’t exist, even though there’s no vertical asymptote
- Dirichlet function: \(D(x)=\begin{cases}1 &x\in\Q\\0 &x\not\in\Q\end{cases}\)
- \(\forall a \in \R, \lim_{x\to a}{D(x)}\) doesn’t exist
- This function is discontinuous at every point.
- \(\forall a \in \R, \lim_{x\to a}{D(x)}\) doesn’t exist
Finding discontinuous points¶
Polynomials:¶
- \(f(x)=\frac{x^2+3x}{x^2+x-6}\)
- Candidates for discontinuity:
- \(\begin{aligned}&x^2+x-6 = 0\\(&x+3.(x-2.=0\\\Rightarrow &x=-3,2\end{aligned}\)
- \(\text{At }x=2: \lim_{x\to 2}{f(x)}=\lim_{x\to 2}{\frac{x}{x-2}=\pm\infty}\Rightarrow\text{ Type II discontinuity @ }x=2.\)
- \(\text{At }x=-3: \lim_{x\to -3}{f(x)}=\lim_{x\to -3}{\frac{x}{x-2}=\frac35}\)
BUT, \(f(-3)\) is undefined \(\Rightarrow\text{ Removable discontinuity}\)
- \(\begin{aligned}&x^2+x-6 = 0\\(&x+3.(x-2.=0\\\Rightarrow &x=-3,2\end{aligned}\)
- Candidates for discontinuity:
Piecewise Functions¶
- \(f(x)=\begin{cases}2x+\frac{6}{2-x} &:x\lt 4 \\8-\frac8x &:x\ge 4\end{cases}\)
- Candidates for discontinuity:
- x=2,
x=0- Vanishing denominators
- But, \(x=0\) is not a candidate, because the expression(s) which is discontinuous at \(x=0\) (\(8-\frac8x\)) is not the definition of \(f\) when \(x=0\), only for \(x\ge4\).
- x=4
- Because there’s a potential for a mismatch of the functions
- x=2,
- \(x=2\)
- \(\lim_{x\to 2}{f(x)}=\lim_{x\to 2}{2(2.+{6\over2-2}=\pm\infty}\)
- Type II Discontinuity
- \(\lim_{x\to 2}{f(x)}=\lim_{x\to 2}{2(2.+{6\over2-2}=\pm\infty}\)
- \(x=4\)
- We need to check both sides
- \(\lim_{x\to 4^-}{f(x)=\lim_{x\to 4}{2x+\frac{6}{2-x}}=8+{6\over-2}=8-3=5}\)
- \(\lim_{x\to 4^+}{f(x)=\lim_{x\to 4}{8-\frac8x}=8-{8\over4}=6}\)
- Type I Discontinuity
- Candidates for discontinuity:
Exponents in the denominator containing \(x\).¶
- \(f(x)=\frac6{2+3^{(\frac4{5-x})}}\)
- Candidates for discontinuity
- \(x=5\)
- \(3^{(\frac4{5-x})}=-2\)
- Not possible, a positive number to any power will be > 0
- \(x=5\)
- \(\lim_{x\to5}{f(x)}\)
- The denominator of the exponent is going to zero, which means the exponent will be either \(\pm\infty\). Because of this, we need to check separately.
- \(\begin{aligned}\\\lim_{x\to5^+}{f(x)}&=\lim_{x\to5^+}{\frac6{2+3^{(\frac4{5-x})}}}\\&=\frac6{2+3^{(\lim_{x\to5^+}{\frac4{5-x}})}}\\&=\frac6{2+3^{-\infty}}\\&=3\end{aligned}\)
- \(\begin{aligned}\\\lim_{x\to5^-}{f(x)}&=\lim_{x\to5^-}{\frac6{2+3^{(\frac4{5-x})}}}\\&=\frac6{2+3^{(\lim_{x\to5^-}{\frac4{5-x}})}}\\&=\frac6{2+3^{+\infty}}\\&=\frac6\infty\\&=0\end{aligned}\)
- Jump Discontinuity
- \(\lim_{x\to5}{f(x)}\)
- \(f\) has a Type I discontinuity at \(x=5\).
- Candidates for discontinuity
May 1 Tirgul. Links to the board and the recording
Forming a continuous function using known continuous functions¶
Theorem 25¶
Theorem
- If \(f\) and \(g\) are continuous at \(x=a\), then the following functions are also continuous at \(x=a\):
- \(f+g\)
- \(f-g\)
- \(f\cdot g\)
- \(\frac fg\mid g(a)\not=0\)
- If \(g\) is continuous at \(x=a\) and \(f\) is continuous at \(g(a)\), then \((f\of g)(x)\) is continuous at \(a\).
Definition: An elementary function is one obtained by adding / subtracting / multiplying / dividing / composing any of the following functions of x:
- \(a^x\)
- \(log_ax\)
- \(sin/cos/tan\ (x)\)
- \(x^n\)
- \(\sqrt[n]{x}\)
- constants
- All the functions we’ve seen except abs. value function, whole number function, Dirichlet function
Theorem 26¶
Theorem
Every elementary function is continuous at every point in it’s domain
Theorem 27¶
Theorem
- If f is continuous at x=a and \(f(a) > 0\), then there exista a neighborhood of a in which \(f(x)>0 \mid\forall x\)
- If f is continuous at x=a and \(f(a) < 0\), then there exista a neighborhood of a in which \(f(x)<0 \mid\forall x\)
Definition: \(f\) is continuous in the closed interval \([a,b]\) if \(f\) is continuous at every point in \((a, b)\), continuous from the right at \(x=a\), and continuous from the left at \(x=b\).
Weirstrass Theorem (28-29)¶
Theorem
- If \(f\) is continuous in \([a,b]\) then \(f\) is bounded there
- i.e. \(\exists m,M \text{ such that } m \le f(x) \le M \mid\forall x\in[a,b]\)
-
If \(f\) is continuous in \([a,b]\) then \(f\) attains it’s maximum and minimum there
- i.e. \(\exists c,d\in[a,b] \text{ such that } f(c) \le f(x) \le f(d) \mid\forall x\in[a,b]\)
- This means that there is a defined upper and lower value of \(f\)
Intermediate Value Theorem (IVT):¶
Theorem
If \(f\) is continuous on a closed interval \([a,b]\) and \(f(a)\not=f(b)\) then for every \(\alpha\) between \(f(a)\) and \(f(b)\) there exists \(c\in[a,b]\) such that \(f(c)=\alpha\).
Specifically: If \(f(a)>0\text{ and }f(b) < 0 \text { then }\exists c\in(a,b) \text{ such that } f(c)=0\)
May 3 Lecture. Omitted Intentionally. Links to the board and the recording.