Differentiability implies continuity

$\newenvironment {prompt}{}{} \newcommand {\todo }[0]{} \newcommand {\oiint }[0]{{\large \bigcirc }\kern -1.56em\iint } \newcommand {\mooculus }[0]{\textsf {\textbf {MOOC}\textnormal {\textsf {ULUS}}}} \newcommand {\npnoround }[0]{\nprounddigits {-1}} \newcommand {\npnoroundexp }[0]{\nproundexpdigits {-1}} \newcommand {\npunitcommand }[1]{\ensuremath {\mathrm {#1}}} \newcommand {\RR }[0]{\mathbb R} \newcommand {\R }[0]{\mathbb R} \newcommand {\N }[0]{\mathbb N} \newcommand {\Z }[0]{\mathbb Z} \newcommand {\sagemath }[0]{\textsf {SageMath}} \newcommand {\d }[0]{\mathop {}\!d} \newcommand {\l }[0]{\ell } \newcommand {\ddx }[0]{\frac {d}{\d x}} \newcommand {\zeroOverZero }[0]{\ensuremath {\boldsymbol {\tfrac {0}{0}}}} \newcommand {\inftyOverInfty }[0]{\ensuremath {\boldsymbol {\tfrac {\infty }{\infty }}}} \newcommand {\zeroOverInfty }[0]{\ensuremath {\boldsymbol {\tfrac {0}{\infty }}}} \newcommand {\zeroTimesInfty }[0]{\ensuremath {\small \boldsymbol {0\cdot \infty }}} \newcommand {\inftyMinusInfty }[0]{\ensuremath {\small \boldsymbol {\infty -\infty }}} \newcommand {\oneToInfty }[0]{\ensuremath {\boldsymbol {1^\infty }}} \newcommand {\zeroToZero }[0]{\ensuremath {\boldsymbol {0^0}}} \newcommand {\inftyToZero }[0]{\ensuremath {\boldsymbol {\infty ^0}}} \newcommand {\numOverZero }[0]{\ensuremath {\boldsymbol {\tfrac {\#}{0}}}} \newcommand {\dfn }[0]{\textbf } \newcommand {\unit }[0]{\mathop {}\!\mathrm } \newcommand {\eval }[1]{\bigg [ #1 \bigg ]} \newcommand {\seq }[1]{\left ( #1 \right )} \newcommand {\epsilon }[0]{\varepsilon } \newcommand {\phi }[0]{\varphi } \newcommand {\iff }[0]{\Leftrightarrow } \DeclareMathOperator {\arccot }{arccot} \DeclareMathOperator {\arcsec }{arcsec} \DeclareMathOperator {\arccsc }{arccsc} \DeclareMathOperator {\si }{Si} \DeclareMathOperator {\scal }{scal} \DeclareMathOperator {\sign }{sign} \newcommand {\arrowvec }[1]{{\overset {\rightharpoonup }{#1}}} \newcommand {\vec }[1]{{\overset {\boldsymbol {\rightharpoonup }}{\mathbf {#1}}}} \DeclareMathOperator {\proj }{\vec {proj}} \newcommand {\veci }[0]{{\boldsymbol {\hat {\imath }}}} \newcommand {\vecj }[0]{{\boldsymbol {\hat {\jmath }}}} \newcommand {\veck }[0]{{\boldsymbol {\hat {k}}}} \newcommand {\vecl }[0]{\vec {\boldsymbol {\l }}} \newcommand {\uvec }[1]{\mathbf {\hat {#1}}} \newcommand {\utan }[0]{\mathbf {\hat {t}}} \newcommand {\unormal }[0]{\mathbf {\hat {n}}} \newcommand {\ubinormal }[0]{\mathbf {\hat {b}}} \newcommand {\dotp }[0]{\bullet } \newcommand {\cross }[0]{\boldsymbol \times } \newcommand {\grad }[0]{\boldsymbol \nabla } \newcommand {\divergence }[0]{\grad \dotp } \newcommand {\curl }[0]{\grad \cross } \newcommand {\lto }[0]{\mathop {\longrightarrow \,}\limits } \newcommand {\bar }[0]{\overline } \newcommand {\surfaceColor }[0]{violet} \newcommand {\surfaceColorTwo }[0]{redyellow} \newcommand {\sliceColor }[0]{greenyellow} \newcommand {\vector }[1]{\left \langle #1\right \rangle } \newcommand {\sectionOutcomes }[0]{} \newcommand {\HyperFirstAtBeginDocument }[0]{\AtBeginDocument } \newcommand {\epstopdfsetup }[0]{\setkeys {ETE}} changed the theorem header of amsthm failed\MessageBreak }}} \newcommand {\GetTitleStringSetup }[0]{\setkeys {gettitlestring}} \newcommand {\Sectionformat }[2]{#1}$

We see that if a function is differentiable at a point, then it must be continuous at that point.

There are connections between continuity and differentiability.

Differentiability Implies Continuity If $f$ is a differentiable function at $x = a$ , then $f$ is continuous at $x=a$ .

To explain why this is true, we are going to use the following definition of the derivative $f'(a) = \lim _{x\to a} \frac {f(x)-f(a)}{x-a}.$

Assuming that $f'(a)$ exists, we want to show that $f(x)$ is continuous at $x=a$ , hence we must show that $\lim _{x\to a} f(x) = f(a).$ Starting with $\lim _{x\to a} \left (f(x) - f(a)\right )$ we multiply and divide by $(x-a)$ to get

$\begin{align*} &= \lim_{x\to a} \left((x-a)\frac{f(x) - f(a)}{x-a}\right) \\ &= \left(\lim_{x\to a} (x-a) \right) \left(\lim_{x\to a}\frac{f(x) - f(a)}{x-a}\right) &\text{Limit Law.} \\ &= \answer[given]{0}\cdot f'(a) = \answer[given]{0}. \end{align*}$ Since $\lim _{x\to a}\left (f(x) - f(a)\right ) = 0$ we see that $\lim _{x\to a} f(x) = f(a)$ , and so $f$ is continuous at $x=a$ .

This theorem is often written as its contrapositive:

If $f(x)$ is not continuous at $x=a$ , then $f(x)$ is not differentiable at $x=a$ .

Thus from the theorem above, we see that all differentiable functions on $\RR$ are continuous on $\RR$ . Nevertheless there are continuous functions on $\RR$ that are not differentiable on $\RR$ .

Which of the following functions are continuous but not differentiable on $\RR$ ?

Consider $f(x) = \begin {cases} x^2 &\text {if $x<3$,}\\ mx+b &\text {if $x\ge 3$.} \end {cases}$ What values of $m$ and $b$ make $f$ differentiable at $x=3$ ?

To start, we know that we must make $f$ both continuous and differentiable. We will start by showing $f$ is continuous at $x=3$ . Write with me: $\begin{align*} \lim_{x\to 3^-} f(x) &= \answer[given]{9}\\ \lim_{x\to 3^+} f(x) &= \answer[given]{m 3 + b}\\ f(3) &= \answer[given]{m 3 + b} \end{align*}$ So for the function to be continuous, we must have $m\cdot 3 + b =9.$ We also must ensure that the value of the derivatives of both pieces of $f$ agree at $x=3$ . Write with me $\begin{align*} \ddx x^2 &= \lim_{h\to 0}\frac{(x+h)^2-x^2}{h}\\ &= \lim_{h\to 0}\frac{x^2 + 2xh + h^2 - x^2}{h}\\ &= \lim_{h\to 0}\frac{2xh + h^2}{h}\\ &= \lim_{h\to 0}\left(2x + h\right)\\ &=2x. \end{align*}$ Moreover, by the definition of a tangent line $\ddx (mx+ b) = \answer [given]{m}$ Hence we must have $\begin{align*} \eval{\ddx x^2}_{x=3} &= \eval{\ddx(mx+b)}_{x=3}\\ \eval{2x}_{x=3} &= \eval{m}_{x=3}\\ 6 &= m. \end{align*}$ Ah! So now $\begin{align*} 9 &= m\cdot 3 + b\\ 9 &= 6\cdot 3 + b\\ 9 &= 18 + b, \end{align*}$ so $b=\answer [given]{-9}$ . Thus setting $m=\answer [given]{6}$ and $b=\answer [given]{-9}$ will give us a differentiable (and hence continuous) piecewise function. We can confirm our results by looking at the graph of $y=f(x)$ : $\graph [xmin=0,xmax=6,ymin=-1,ymax=15]{y=x^2\left \{x<3\right \},y=mx+b\left \{3 \leq x\right \},m=6,b=-9}$

Press...	...to do
left/right arrows	Move cursor
shift+left/right arrows	Select region
ctrl+a	Select all
ctrl+x/c/v	Cut/copy/paste
ctrl+z/y	Undo/redo
ctrl+left/right	Add entry to list or column to matrix
shift+ctrl+left/right	Add copy of current entry/column to to list/matrix
ctrl+up/down	Add row to matrix
shift+ctrl+up/down	Add copy of current row to matrix
ctrl+backspace	Delete current entry in list or column in matrix
ctrl+shift+backspace	Delete current row in matrix

Type...	...to get
norm	$\|\|\blue{[?]}\|\|$
text	$\text{\blue{[?]}}$
sym_name	$\backslash\texttt{\blue{[?]}}$
abs	$\left\|\blue{[?]}\right\|$
sqrt	$\sqrt{\blue{[?]}}$
paren	$\left(\blue{[?]}\right)$
floor	$\lfloor \blue{[?]} \rfloor$
factorial	$\blue{[?]}!$
exp	${\blue{[?]}}^{\blue{[?]}}$
sub	${\blue{[?]}}_{\blue{[?]}}$
frac	$\dfrac{\blue{[?]}}{\blue{[?]}}$
int	$\displaystyle\int{\blue{[?]}}d\blue{[?]}$
defi	$\displaystyle\int_{\blue{[?]}}^{\blue{[?]}}\blue{[?]}d\blue{[?]}$
deriv	$\displaystyle\frac{d}{d\blue{[?]}}\blue{[?]}$
sum	$\displaystyle\sum_{\blue{[?]}}^{\blue{[?]}}\blue{[?]}$
prod	$\displaystyle\prod_{\blue{[?]}}^{\blue{[?]}}\blue{[?]}$
root	$\sqrt[\blue{[?]}]{\blue{[?]}}$
vec	$\left\langle \blue{[?]} \right\rangle$
mat	$\left(\begin{matrix} \blue{[?]} \end{matrix}\right)$
*	$\cdot$
infinity	$\infty$
arcsin	$\arcsin\left(\blue{[?]}\right)$
arccos	$\arccos\left(\blue{[?]}\right)$
arctan	$\arctan\left(\blue{[?]}\right)$
sin	$\sin\left(\blue{[?]}\right)$
cos	$\cos\left(\blue{[?]}\right)$
tan	$\tan\left(\blue{[?]}\right)$
sec	$\sec\left(\blue{[?]}\right)$
csc	$\csc\left(\blue{[?]}\right)$
cot	$\cot\left(\blue{[?]}\right)$
log	$\log\left(\blue{[?]}\right)$
ln	$\ln\left(\blue{[?]}\right)$
alpha	$\alpha$
beta	$\beta$
gamma	$\gamma$
delta	$\delta$
epsilon	$\epsilon$
zeta	$\zeta$
eta	$\eta$
theta	$\theta$
iota	$\iota$
kappa	$\kappa$
lambda	$\lambda$
mu	$\mu$
nu	$\nu$
xi	$\xi$
omicron	$\omicron$
pi	$\pi$
rho	$\rho$
sigma	$\sigma$
tau	$\tau$
upsilon	$\upsilon$
phi	$\phi$
chi	$\chi$
psi	$\psi$
omega	$\omega$
Gamma	$\Gamma$
Delta	$\Delta$
Theta	$\Theta$
Lambda	$\Lambda$
Xi	$\Xi$
Pi	$\Pi$
Sigma	$\Sigma$
Phi	$\Phi$
Psi	$\Psi$
Omega	$\Omega$

Controls

Symbols

Settings