Functions, limits and continuity – university Year 1 maths lesson to download as PDF

Example 1

Prove \(\lim_{x\to2}(3x-1)=5\).

Given \(\varepsilon>0\) choose \(\delta=\varepsilon/3\). If \(0<|x-2|\lt \delta\) then \(|(3x-1)-5|=3|x-2|<3\delta=\varepsilon\).

Example 2

Prove \(\lim_{x\to1}x^2=1\).

\(|x^2-1|=|x-1||x+1|\). Restrict to \(|x-1|<1\), so \(0\lt x<2\) and \(|x+1|<3\). Then \(|x^2-1|<3|x-1|\). Choose \(\delta=\min(1,\varepsilon/3)\); then \(|x^2-1|<3\delta\le\varepsilon\).

Example 3

Find \(\lim_{x\to0}x^2\sin(1/x)\).

\(-x^2\le x^2\sin(1/x)\le x^2\), and both bounds tend to 0, so by the squeeze theorem the limit is 0.

Example 4: removable discontinuity

Let \(f(x)=\dfrac{x^2-9}{x-3}\) for \(x\ne3\). Is the discontinuity at 3 removable?

Factorising gives \(f(x)=x+3\) for \(x\ne3\), so the limit at 3 is 6. Defining \(f(3)=6\) makes the function continuous, so the discontinuity is removable. Simplify first; never substitute \(x=3\) into the original fraction.

Example 5

Locate the root of \(f(x)=x^3-2x-5\).

\(f(2)=-1\) and \(f(3)=16\). The midpoint 2.5 gives \(f(2.5)=5.625>0\), so the root is in \((2,2.5)\). Next \(f(2.25)\approx1.89>0\), so it is in \((2,2.25)\). Ten steps give the root to within about 0.001.

Worked example 6: A limit at infinity with square roots

Evaluate \(\displaystyle\lim_{x\to\infty}\left(\sqrt{x^2+3x}-x\right)\).

The form is \(\infty-\infty\), so multiply and divide by the conjugate: \(\dfrac{(x^2+3x)-x^2}{\sqrt{x^2+3x}+x}=\dfrac{3x}{\sqrt{x^2+3x}+x}\). Divide numerator and denominator by \(x\) (which is positive): \(\dfrac3{\sqrt{1+3/x}+1}\). As \(x\to\infty\) this tends to \(\dfrac32\). A common mistake is to say the limit is \(0\) because "\(\sqrt{x^2}=x\)"; the correction term \(3x\) is not negligible.

Worked example 7: An epsilon-delta proof with a quotient

Prove that \(\displaystyle\lim_{x\to2}\dfrac1x=\dfrac12\).

Rough work: \(\left|\dfrac1x-\dfrac12\right|=\dfrac{|x-2|}{2|x|}\). To keep \(|x|\) away from zero, restrict to \(|x-2|<1\), so \(1\lt x<3\) and \(\dfrac1{2|x|}\lt \dfrac12\). Proof: given \(\varepsilon>0\) let \(\delta=\min(1,2\varepsilon)\). If \(0<|x-2|\lt \delta\), then \(\left|\dfrac1x-\dfrac12\right|\lt \dfrac12|x-2|\lt \dfrac12\cdot2\varepsilon=\varepsilon\). As required. The idea is always the same: bound the awkward factor first, then choose \(\delta\) as a minimum.

Worked example 8: The squeeze theorem with a trigonometric term

Evaluate \(\displaystyle\lim_{x\to\infty}\dfrac{\sin x}{x}\).

Because \(-1\le\sin x\le1\) and \(x>0\) we have \(-\dfrac1x\le\dfrac{\sin x}x\le\dfrac1x\). Both outer functions tend to \(0\) as \(x\to\infty\), so the squeeze theorem gives a limit of \(0\). Notice that \(\sin x\) itself has no limit, but dividing by a quantity that grows without bound tames the oscillation completely.

Worked example 9: Choosing a constant for continuity

Find all values of \(a\) and \(b\) so that \(f(x)=\begin{cases}ax+1&x<0\\ b&x=0\\ \dfrac{\sin ax}{x}&x>0\end{cases}\) is continuous at \(0\).

The left limit is \(a\cdot0+1=1\). The right limit is \(\dfrac{\sin ax}x=a\cdot\dfrac{\sin ax}{ax}\to a\). Continuity needs left limit \(=\) right limit \(=f(0)\), so \(a=1\) and \(b=1\). Any other choice of \(a\) leaves a jump, and any other choice of \(b\) leaves a removable discontinuity.

Worked example 10: A fixed point of a continuous function

Prove that every continuous function \(f:[0,1]\to[0,1]\) has a fixed point, that is a number \(c\) with \(f(c)=c\).

Let \(g(x)=f(x)-x\), which is continuous on \([0,1]\). Then \(g(0)=f(0)\ge0\) because \(f(0)\in[0,1]\), and \(g(1)=f(1)-1\le0\) because \(f(1)\le1\). If either value is zero, we have found the fixed point. Otherwise \(g(0)>0>g(1)\), and the intermediate value theorem gives \(c\in(0,1)\) with \(g(c)=0\), that is \(f(c)=c\). Geometrically the graph of \(f\) must cross the diagonal.

Worked example 11: Bisection to a stated accuracy

How many bisection steps are needed to locate a root of a continuous function in \([2,3]\) to within \(10^{-4}\)?

After \(k\) steps the interval has length \(\dfrac1{2^k}\), so the midpoint is within \(\dfrac1{2^{k+1}}\) of the root. We need \(\dfrac1{2^{k+1}}<10^{-4}\), that is \(2^{k+1}>10\,000\). Since \(2^{13}=8192\) and \(2^{14}=16\,384\), we need \(k+1=14\), so \(k=13\) steps. Bisection is slow but certain: each step gains exactly one binary digit.

Worked example 12: Limits of a piecewise function at a boundary

Let \(f(x)=\dfrac{x^2-x-6}{x-3}\) for \(x\ne3\). Find \(\displaystyle\lim_{x\to3}f(x)\), decide whether \(f\) can be made continuous at \(3\), and state the natural domain.

Factorise the numerator: \(x^2-x-6=(x-3)(x+2)\). For \(x\ne3\) the fraction simplifies to \(x+2\), so the limit as \(x\to3\) is \(5\). If we define \(f(3)=5\) the new function is continuous everywhere, so the break at \(3\) was removable. The natural domain of the original formula is every real number except \(3\), because the denominator vanishes there. It is the simplification, not the substitution, that reveals the true behaviour of the function near the point.

Worked example 13: Showing that a function is not continuous on an interval

Why does the intermediate value theorem not guarantee a root of \(f(x)=\dfrac{x}{|x|}\) on \([-1,1]\), even though \(f(-1)<0\lt f(1)\)?

The function takes only the values \(-1\) and \(1\) for \(x\ne0\), and is not defined at \(0\). It has a jump discontinuity at \(0\), so it is not continuous on the closed interval \([-1,1]\). The theorem requires continuity at every point of the interval, and without it the graph can leap from below the axis to above it without ever crossing. This example shows that the hypotheses are not decoration: remove one and the conclusion fails.

Worked example 14: Using standard limits

Evaluate \(\displaystyle\lim_{x\to0}\dfrac{e^{3x}-1}{x}\) and \(\displaystyle\lim_{x\to0}\dfrac{1-\cos4x}{x^2}\).

For the first limit write \(\dfrac{e^{3x}-1}x=3\cdot\dfrac{e^{3x}-1}{3x}\). With \(u=3x\to0\) the standard limit \(\dfrac{e^u-1}u\to1\) gives the answer \(3\). For the second, \(\dfrac{1-\cos4x}{x^2}=16\cdot\dfrac{1-\cos u}{u^2}\) with \(u=4x\), and the standard limit \(\dfrac12\) gives \(16\times\dfrac12=8\). In both cases the trick is to make the argument of the function match the denominator.