\(x^2\) \(\lfloor x \rfloor \) \(|x|\)
\(\frac {\sin (x)}{x}\)
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 apply the Difference Law to the left hand side \[ \lim _{x\to a}f(x) - \lim _{x\to a}f(a) = 0 , \]
and use continuity of a
constant to obtain that \[ \lim _{x\to a}f(x) - f(a) = 0 . \]
Next, we add \(f(a)\) on both sides and get that \[ \lim _{x\to a}f(x) = f(a). \]
Now 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 \(x=a\).
From the theorem above, we see that all differentiable functions on \((-\infty , \infty )\) are continuous on \((-\infty , \infty )\). Nevertheless there are continuous functions on \((-\infty , \infty )\) that are not differentiable on \((-\infty , \infty )\).
Which of the following functions are continuous but not differentiable on \((-\infty , \infty )\)?
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 \(f\)
must be continuous at \(x=3\), since it has to be differentiable there. We will start by
making \(f\) 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
function is differentiable at \(x=3\). In other words, we have to ensure that the following
limit exists \[ \lim _{x\to 3}\frac {f(x)-f(3)}{x-3}.\]
In order to compute this limit, we have to compute the two one-sided limits
\[ \lim _{x\to 3^{+}}\frac {f(x)-f(3)}{x-3} \]
and \[ \lim _{x\to 3^{-}}\frac {f(x)-f(3)}{x-3}, \]
since \(f(x)\) changes expression at \(x=3\). Write with me \begin{align*} \lim _{x\to 3^{-}}\frac {f(x)-f(3)}{x-3}&= \lim _{x\to 3}\frac {x^2 -9}{x-3}\\ &= \lim _{x\to 3^{-}}\frac {\cancel {(x-3)}(x+3)}{\cancel {x-3}}\\ &= \lim _{x\to 3^{-}}\left (x+3\right )\\ &=6, \end{align*}
and
\begin{align*} \lim _{x\to 3^{+}}\frac {f(x)-f(3)}{x-3}&= \lim _{x\to 3}\frac {mx+b -(3m+b)}{x-3}\\ &= \lim _{x\to 3^{+}}\frac {mx-3m}{x-3}\\ &= \lim _{x\to 3^{+}}\frac {m\cancel {(x-3)}}{\cancel {x-3}}\\ &= \lim _{x\to 3^{+}}m\\ &=m. \end{align*}
Hence, we must have
\[ m=6.\]
Ah! So now the equation that must be satisfied \begin{align*} 9 &= m\cdot 3 + b,\\ becomes\hspace {0.3in} 9 &= 6\cdot 3 + b.\\ \end{align*}
Therefore, \(b=\answer [given]{-9}\). Thus setting \(m=\answer [given]{6}\) and \(b=\answer [given]{-9}\) will give us a function that is differentiable (and hence continuous) at \(x=3\).
Can we tell from its graph whether the function is differentiable or not at a point \(a\)?
What does the graph of a function \(f\) possibly look like when \(f\) is not differentiable at
\(a\)?
The function in figure A is not continuous at \(a\), and, therefore, it is not differentiable
there.
Each of the figures A-D depicts a function that is not differentiable at \(a=1\).
![[Picture]](digInDifferentiabilityImpliesContinuity.online-bf3be024e2c50a742b40462acf906bbb.svg) |
![[Picture]](digInDifferentiabilityImpliesContinuity.online-d8ac6ad38f5e0eca9197a07e62fa9291.svg) |
![[Picture]](digInDifferentiabilityImpliesContinuity.online-35e76a25d2abf8d9e861d24eed979072.svg) |
![[Picture]](digInDifferentiabilityImpliesContinuity.online-833458e06a4edd977fc5939c7fdb55f0.svg) |
In figures \(B\)–\(D\) the functions are continuous at \(a\), but in each case the limit
\[ \lim _{x\to a} \frac {f(x)-f(a)}{x-a} \]
does not
exist, for a different reason.
In figure \(B\)
\[ \lim _{x\to a^{+}} \frac {f(x)-f(a)}{x-a}\ne \lim _{x\to a^{-}} \frac {f(x)-f(a)}{x-a}. \]
In figure \(C\)
\[ \lim _{x\to a} \frac {f(x)-f(a)}{x-a}=\infty . \]
In figure \(D\) the two one-sided limits don’t exist and neither one of them is
infinity.
So, if at the point \(a\) a function either has a "jump" in the graph, or a corner, or what looks like a “vertical tangent line", or if it rapidly oscillates near \(a\), then the function is not differentiable at \(a\).