Solved Examples

This section works through twelve fully solved examples and one guided activity, applying every idea from 8.1.1–8.1.11 to concrete functions. Several examples also introduce standard limit results that are quoted, without re-derivation, throughout the harder examples: $\displaystyle\lim_{x\to0}\frac{e^x-1}{x}=1$, $\displaystyle\lim_{x\to0}\frac{a^x-1}{x}=\log a$, $\displaystyle\lim_{x\to0}(1+t)^{1/t}=e$, $\displaystyle\lim_{x\to0}(1-t)^{1/t}=e^{-1}=\frac1e$, $\displaystyle\lim_{x\to0}\frac{\log(1+x)}{x}=1$, and $\displaystyle\lim_{x\to0}\frac{\log(1-x)}{x}=-1$ (these can be proved using L'Hospital's rule or power-series expansions, studied at a more advanced stage) — together with the already-familiar $\displaystyle\lim_{\theta\to0}\frac{\sin\theta}{\theta}=1=\lim_{\theta\to0}\frac{\tan\theta}{\theta}$.

Example 1. Discuss the continuity of $f(x)=|x-3|$ at $x=3$ (Fig. 8.8). By the definition of the modulus function, $f(x)=-(x-3)$ for $x<3$ and $f(x)=x-3$ for $x\ge3$. Here $f(3)=3-3=0$; also $\displaystyle\lim_{x\to3^-} f(x)=-(3-3)=0$ and $\displaystyle\lim_{x\to3^+} f(x)=3-3=0$, so both one-sided limits equal $0=\lim_{x\to3}f(x)$, and $\displaystyle\lim_{x\to3} f(x)=f(3)=0$. Therefore $f(x)$ is continuous at $x=3$.

Example 2. Determine whether $f$ is continuous on $\mathbb{R}$, where $f(x)=3x+1$ for $x<2$, $f(x)=7$ for $2\le x<4$, and $f(x)=x^2-8$ for $x\ge4$; if discontinuous, state the type. Since the function is built from polynomials on each piece, any discontinuity can only occur where the definition switches, i.e. at $x=2$ and $x=4$. At $x=2$: $f(2)=7$ (given), $\displaystyle\lim_{x\to2^-} f(x)=\lim_{x\to2}(3x+1)=7$, and $\displaystyle\lim_{x\to2^+} f(x)=7$; so $\displaystyle\lim_{x\to2} f(x)=f(2)=7$, and $f$ is continuous at $x=2$. At $x=4$: $f(4)=4^2-8=8$; $\displaystyle\lim_{x\to4^-} f(x)=\lim_{x\to4}(7)=7$, while $\displaystyle\lim_{x\to4^+} f(x)=\lim_{x\to4}(x^2-8)=16-8=8$. Since $7\ne8$, $\displaystyle\lim_{x\to4} f(x)$ does not exist, so $f$ is discontinuous at $x=4$, with a jump discontinuity there (both one-sided limits exist but differ); $f$ is continuous everywhere else on $\mathbb{R}$.

Example 3. Test whether $f(x)=\dfrac{x^2+16x+48}{x+4}$ for $x\ne-4$, $f(-4)=8$, is continuous at $x=-4$. Here $f(-4)=8$ is defined. Factorising, $x^2+16x+48=(x+4)(x+12)$, so for $x\ne-4$,

So $\displaystyle\lim_{x\to-4} f(x)=f(-4)=8$, and by definition $f(x)$ is continuous at $x=-4$.

Example 4. Discuss the continuity of $f(x)=\sqrt{9-x^2}$ on the interval $[-3,3]$. The domain of $f$ is exactly $[-3,3]$, since $f$ needs $9-x^2\ge0$. For any $a\in(-3,3)$, $\displaystyle\lim_{x\to a} f(x)=\lim_{x\to a}\sqrt{9-x^2}=\sqrt{9-a^2}=f(a)$, so $f$ is continuous at every interior point. At the endpoints, $f(3)=0$ and $f(-3)=0$; also $\displaystyle\lim_{x\to3^-} f(x)=f(3)=0$ and $\displaystyle\lim_{x\to-3^+} f(x)=f(-3)=0$. So $f$ is continuous on $(-3,3)$, continuous to the right at $x=-3$, and continuous to the left at $x=3$; hence $f(x)=\sqrt{9-x^2}$ is continuous on the closed interval $[-3,3]$.

Example 5. Show that $f(x)=\lfloor x\rfloor$ is not continuous at $x=0,1$ on the interval $[-1,2)$ (Fig. 8.9): $f(x)=-1$ for $x\in[-1,0)$, $f(x)=0$ for $x\in[0,1)$, and $f(x)=1$ for $x\in[1,2)$. At $x=0$: $f(0)=0$, but $\displaystyle\lim_{x\to0^-} f(x)=-1$ while $\displaystyle\lim_{x\to0^+} f(x)=0$; since these differ, $f$ is discontinuous at $x=0$. At $x=1$: $f(1)=1$, but $\displaystyle\lim_{x\to1^-} f(x)=0$ while $\displaystyle\lim_{x\to1^+} f(x)=1$; since these differ too, $f$ is discontinuous at $x=1$. Hence $f(x)=\lfloor x\rfloor$ is not continuous at $x=0$ or $x=1$ within $[-1,2)$ (each is a jump discontinuity, visible as the two steps in Fig. 8.9).

Example 6. Discuss the continuity of $f(x)=x^2\sin\!\left(\dfrac1x\right)$ for $x\ne0$, $f(0)=0$, at $x=0$. Since $-1\le\sin\!\left(\dfrac1x\right)\le1$ for every $x\ne0$, multiplying throughout by $x^2\ (\ge0)$ gives $-x^2\le x^2\sin\!\left(\dfrac1x\right)\le x^2$. Taking the limit as $x\to0$ throughout, both outer bounds $-x^2$ and $x^2$ go to $0$, so by the Squeeze Theorem, $\displaystyle\lim_{x\to0} x^2\sin\!\left(\dfrac1x\right)=0$, i.e. $\displaystyle\lim_{x\to0} f(x)=0=f(0)$. Hence $f(x)$ is continuous at $x=0$.

Example 7. Find $k$ if $f(x)=\dfrac{xe^x+\tan x}{\sin 3x}$ for $x\ne0$, $f(0)=k$, is continuous at $x=0$. Since $f$ must equal its own limit at $0$,

using $\lim_{x\to0}\frac{\sin\theta}{\theta}=1=\lim_{x\to0}\frac{\tan\theta}{\theta}$ as $x\to0,\,3x\to0$. So $k=\dfrac23$.

Example 8. If $f$ is continuous at $x=1$, where $f(x)=\dfrac{\sin(\pi x)}{x-1}+a$ for $x<1$, $f(1)=2\pi$, and $f(x)=\dfrac{1+\cos(\pi x)}{\pi(1-x)^2}+b$ for $x>1$, find $a$ and $b$. Continuity at $x=1$ requires $\displaystyle\lim_{x\to1^-} f(x)=\lim_{x\to1^+} f(x)=f(1)$. For the left limit, put $x-1=t$ so $x=1+t$, $t\to0^-$: $\sin(\pi x)=\sin(\pi+\pi t)=-\sin(\pi t)$, so

For the right limit, put $1-x=\theta$ so $x=1-\theta$, $\theta\to0^+$; using $1+\cos(\pi x)=1-\cos(\pi\theta)=2\sin^2\!\left(\frac{\pi\theta}{2}\right)$,

So $a=3\pi$, $b=\dfrac{3\pi}{2}$.

Example 9. Identify discontinuities for the following as either jump or removable, on $\mathbb{R}$:

(1) $f(x)=\dfrac{x^2-3x-18}{x-6}$: this rational function is continuous for every $x\ne6$, so $f(6)$ is not defined. Factorising, $x^2-3x-18=(x-6)(x+3)$, so $\displaystyle\lim_{x\to6} f(x)=\lim_{x\to6}(x+3)=9$. Since $f(6)$ is undefined but the limit exists, $f$ has a removable discontinuity at $x=6$.

(2) $g(x)=3x+1$ for $x<3$, $g(x)=2-3x$ for $x\ge3$: this is built from two different polynomials meeting at $x=3$. Here $g(3)=2-3(3)=-7$; $\displaystyle\lim_{x\to3^-} g(x)=\lim_{x\to3}(3x+1)=10$, while $\displaystyle\lim_{x\to3^+} g(x)=\lim_{x\to3}(2-3x)=-7$. Since $10\ne-7$, $\displaystyle\lim_{x\to3} g(x)$ does not exist, so $g$ has a jump discontinuity at $x=3$.

(3) $h(x)=13-x^2$ for $x<5$, $h(x)=13-5x$ for $x>5$: $h(5)$ is not defined by either piece. $\displaystyle\lim_{x\to5^-} h(x)=13-25=-12$ and $\displaystyle\lim_{x\to5^+} h(x)=13-25=-12$ agree, so $\displaystyle\lim_{x\to5} h(x)=-12$ exists even though $h(5)$ is undefined — a removable discontinuity at $x=5$.

Note: the standard limits $\lim_{x\to0}\frac{e^x-1}{x}=1$, $\lim_{x\to0}\frac{a^x-1}{x}=\log a$, $\lim_{x\to0}(1+t)^{1/t}=e$, $\lim_{x\to0}(1-t)^{1/t}=e^{-1}=\frac1e$, $\lim_{x\to0}\frac{\log(1+x)}{x}=1$ and $\lim_{x\to0}\frac{\log(1-x)}{x}=-1$ are used from here on without re-proof; they can be established using L'Hospital's rule or a power-series expansion, taken up at a more advanced stage.

Example 10. Show that $f(x)=\dfrac{5^{\cos x}-e^{(\pi/2-x)}}{\cot x}$ for $x\ne\pi/2$, $f(\pi/2)=\log5-e$, has a removable discontinuity at $x=\pi/2$, and redefine it to be continuous there. Here $f(\pi/2)=\log5-e$ is defined. Put $\dfrac\pi2-x=t$, so $x=\dfrac\pi2-t$, $t\to0$ as $x\to\dfrac\pi2$: then $\cos x=\sin t$, $\cot x=\tan t$, so

and dividing every piece by $t$ (valid since $t\ne0$) and using $\sin t\ne0$ near $0$, this reduces, using $\lim_{t\to0}\frac{\sin t}t=1=\lim_{t\to0}\frac{\tan t}t$, $\lim_{t\to0}\frac{e^t-1}t=1$ and $\lim_{t\to0}\frac{a^t-1}t=\log a$, to

Since $f(\pi/2)=\log5-e$ is defined and $\displaystyle\lim_{x\to\pi/2} f(x)=\log5-1$ exists, but $\log5-1\ne\log5-e$, the function has a removable discontinuity at $x=\pi/2$; it is repaired by redefining $f(\pi/2)=\log5-1$, giving the continuous function

Example 11. If $f(x)=\left(\dfrac{3x+2}{2-5x}\right)^{1/x}$ for $x\ne0$ is continuous at $x=0$, find $f(0)$. Since $f$ must equal its limit at $0$, write $\dfrac{3x+2}{2-5x}=\dfrac{2\left(1+\frac{3x}2\right)}{2\left(1-\frac{5x}2\right)}=\dfrac{\left(1+\frac{3x}2\right)^{1/x}}{\left(1-\frac{5x}2\right)^{1/x}}$, so

using $\lim_{x\to0}(1+kx)^{1/(kx)}=e$. So $f(0)=e^4$. …