# 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 $$$\tag{*} 4x^3-6x^2+3x-2=0.$$$
• 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.

# 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 $$\lim_{x\to\infty}t(x)=10$$.