Continuity 3

From limit laws to continuity

  • The limit laws can be used to get continuity statements.
    • Theorem. Let \(f(x)\) and \(g(x)\) be functions, and \(a\) a number. If \(f(x)\) and \(g(x)\) are continuous at \(a\), then the following functions are also continuous at \(a\). \[ f(x)\pm g(x)\quad f(x)g(x)\quad\frac{f(x)}{g(x)}\text{ if }g(a)\ne0. \]
    • Theorem. (a) Any polynomial function is continuous everywhere. (b) Any rational function is continuous wherever it is defined; that is, it is continuous on its domain.
    • The root law implies that for any positive integer \(n\), the function \(f(x)=\sqrt[n]{x}\) is continuous on its domain.
    • Recall that for any positive number \(b\), the function \(f(x)=b^x\) was defined in a way that it is continous.
    • Let \(f(x)\) be a one-to-one function, which is continuous on its domain. Recall that if we mirror the graph \(y=f(x)\) with respect to the line \(x=y\), we get the graph of the inverse function \(y=f^{-1}(x)\). Since a function is continous on its domain precisely when its graph is a continuous curve, this shows that the inverse function \(f^{-1}(x)\) is continuous on its domain.
    • In particular, this implies that the logarithmic functions are continous on their domain

From limit laws to continuity 2

  • Recall that the function \(f(\theta)=(\cos\theta,\sin\theta)\) assigns to the angle \(\theta\) the point \(P(\cos\theta,\sin\theta)\) on the unit circle. Since this traces out the unit circle in a continuous manner, and it's defined everywhere, the functions \(\cos\theta\), \(\sin\theta\) are continuous everywhere.
    • As a consequence of that, the other trigonometric functions are continuous on their domain too.
    • Also, for example, since we have \[ \lim_{x\to\pi/2^-}\tan(x)=\lim_{x\to\pi/2^-}\frac{\sin(x)}{\cos(x)}=\infty, \] the function \(\tan(x)\) has an infinity discontinuity at \(x=\frac{\pi}{2}\).
    • The inverse trigonometric functions are continuous on their domain.
  • An additional way of combining continuous functions to get continuous functions is composition.
    • Theorem. If \(\lim_{x\to a}g(x)=b\) and \(f(x)\) is continuous at \(b\), then \(\lim_{x\to a}f(g(x))=f(b)\). In other words, \[ \lim_{x\to a}f(g(x))=f(\lim_{x\to a}g(x)). \]
    • Theorem. If \(g(x)\) is continuous at \(a\), and \(f(x)\) is continous at \(g(a)\), then the composite \((f\circ g)(x)=f(g(x))\) is continuous at \(a\).
    • Exercises. 2.5: 28, 32, 36, 40, 48

The Intermediate value theorem

  • As promised last week, we now cover the Intermediate value theorem
    • The Intermediate value theorem. Let \(f(x)\) be a function which is continuous on a closed interval \([a,b]\). Suppose that \(f(a)\ne f(b)\). Let \(N\) be a number between \(f(a)\) and \(f(b)\). Then there exists a number \(c\) in the interval \((a,b)\) such that \(f(c)=N\).
    • This theorem can be used to find solutions of equations. For example, consider the equation \[\begin{equation}\tag{*} 4x^3-6x^2+3x-2=0. \end{equation}\]
    • Let \(f(x)=4x^3-6x^2+3x-2\).
    • Note that we have \(f(x=1)=-1\) and \(f(x=2)=4\).
    • Since \(f(x)\) is a polynomial function, it is continuous on the interval \([1,2]\).
    • Let \(N=0\). Since \(f(x=1)<N<f(x=2)\), the Intermediate value theorem implies that there is a \(c\) in \((1,2)\) such that \(f(x=c)=0\).
    • This means that the equation (*) has a solution in the interval \((1,2)\).
    • Exercises. 2.5: 54,56.

A one-to-one function with domain an interval is strictly monotonous

  • To finish, let us show that if \(f(x)\) is a one-to-one function, the domain \(D\) of which is one connected interval (with at least one endpoint possibly \(\pm\infty\)), then \(f(x)\) is strictly monotonous.
    • Suppose for contradiction that this is not true. Up to symmetry, we can suppose that there exist three points \(a<b<c\) in \(D\) such that \(f(a)<f(b)\) and \(f(b)>f(c)\).
    • Up to symmetry, we can suppose moreover that \(f(a)>f(c)\).
    • Since \(D\) is one connected interval, if it contains \(b\) and \(c\), it contains the entire closed interval \([b,c]\).
    • Then the Intermediate value theorem implies that there is a number \(d\) in \((b,c)\) such that \(f(d)=f(a)\).
    • But as \(a<b<d<c\), we have \(a\ne d\) and \(f(a)=f(d)\), which contradicts the assumption that \(f(x)\) is one-to-one.

Limits at infinity

A Philosophical example

  • Achilles and the Tortoise
    • Achilles and the tortoise are having a footrace.
    • Achilles is such a nice heros, he gives the tortoise a headstart of 100 metres.
    • But then the Ancient Greek philosopher Zeno of Elea shows up, and he claims he can't reach the tortoise!
    • First, he has to halve the distance between them.
    • Then he has to halve the distance again.
    • After which he needs to close in by halving the distance again, ...
  • Let's model this
    • Let \(x\) measure the number of times Achilles has halved the distance.
    • Let \(f(x)\) measure the distance itself.
    • That is, we have \(f(x)=2^{-x}\cdot100\).
  • Zeno hasn't heard of limits at infinity
    • (Well, he was about 2000 years early for that.)
    • Look at the graph of \(y=f(x)\). You can just Google y=2^(-x)*100 for this.
    • Although \(f(x)>0\) for all \(x\), we can get \(f(x)\) arbitrarily close to 0 by choosing large enough \(x\).
    • We'll express this by writing \(\lim_{x\to\infty}f(x)=0\).
    • Definitions will follow after our Philosophical excursion has ended.

Zeno has chosen a weird variable, let's fix it

  • Let's describe the situation with a time variable \(t\).
    • (Without any claim to Mythological or Biological precision), let's assume that Achilles is running at a speed of 11 m/s, and the turtle is crawling at a speed of 1 m/s.
    • The the distance of Achilles from the start is \(d_1(t)=11t\), and the distance of the tortoise from the start is \(100+t\).
    • That is, their relative distance is \(f(t)=d_2(t)-d_1(t)=100-10t\)
    • Therefore Achilles overtakes the turtle at the zero of the function, \(t=10\) s.
  • Transformation formula \(t(x)\)
    • The function \(t(x)\) measures the time required for the distance to have been halved \(x\) times.
    • If \(f(x)=100\cdot2^{-x}\) is the portion of the distance that remains to be covered, the portion that has been covered is \(1-f(x)=1-2^{-x}\).
    • Since the starting distance is 100 m/s, the relative distance that has been covered is \(100(1-2^{-x})\).
    • Since the relative speed is 10 m/s, this gives \(t(x)=10(1-2^{-x})\).
  • Transformation formula \(x(t)\)
    • Exercise. Find the domain, the range, and the inverse of \(t(x)=10(1-2^{-x})\).
    • In particular, as the range of \(t(x)\) is \((-\infty,10)\), Zeno only covers the time interval \((-\infty,10)\).
    • Check out the graph of \(t(x)\). If you shot a movie of this with \(x\) as the time variable, time would get slower and slower as Achilles approached the tortoise, and he would never overtake it.
  • Limits
    • Note that we have \(\lim_{t\to10^-}x(t)=\infty\).
    • Correspondingly, we will see that we have \(\lim_{x\to\infty}t(x)=10\).