Continuity and Differentiability — chapter notes

A real function $f$ is continuous at a point $c$ of its domain if

$\lim_{x \to c^-} f(x) = \lim_{x \to c^+} f(x) = f(c)$.

So three numbers must agree: the left-hand limit, the right-hand limit and the value. If the one-sided limits differ (a jump), or the limit exists but differs from $f(c)$, then $f$ is discontinuous at $c$. Example: $f(x) = x + 1$ for $x < 1$ and $4 - x$ for $x \ge 1$ has left-hand limit $2$ and right-hand limit $3$ at $x = 1$: not continuous. A function is continuous if it is continuous at every point of its domain; it is continuous on $[a, b]$ if it is continuous on $(a, b)$, from the right at $a$ and from the left at $b$.

For a function defined by different formulas on different intervals, each piece (usually a polynomial or a standard function) is continuous on its own interval, so only the break points need testing.

Routine: (1) list the break points; (2) at each, find the value from the piece whose inequality includes the point, and the two one-sided limits from the neighbouring pieces; (3) for unknown constants, equate them; (4) state the conclusion in words.

Example: $kx + 1$ ($x \le 2$) and $3x - 1$ ($x > 2$): $2k + 1 = 5$, so $k = 2$. Two constants usually need two break points, giving two equations. A missing point, such as $\dfrac{x^2 - 9}{x - 3}$ at $3$, is filled by defining the value to be the limit ($6$).

Polynomials, $\sin x$, $\cos x$, $e^x$ and $|x|$ are continuous on $\mathbb{R}$; $\log x$ on $(0, \infty)$; $\tan x$ wherever it is defined; rational functions wherever the denominator is non-zero. If $f$ and $g$ are continuous at $c$, so are $f + g$, $f - g$, $fg$ and $\dfrac{f}{g}$ (when $g(c) \ne 0$); compositions of continuous functions are continuous.

The greatest integer function $[x]$ (greatest integer $\le x$) is constant on each interval $[n, n + 1)$ and jumps by $1$ at each integer $n$: left-hand limit $n - 1$, right-hand limit $n$. So $[x]$ is continuous at every non-integer and discontinuous at every integer.

$f$ is differentiable at $c$ if $f'(c) = \lim_{h \to 0}\dfrac{f(c + h) - f(c)}{h}$ exists, that is, the left-hand derivative ($h \to 0^-$) and right-hand derivative ($h \to 0^+$) exist and are equal.

Differentiable $\Rightarrow$ continuous: $f(c + h) - f(c) = \dfrac{f(c + h) - f(c)}{h} \cdot h \to f'(c) \cdot 0 = 0$. The converse is false: $|x|$ is continuous at $0$, but its one-sided derivatives are $-1$ and $1$, so it is not differentiable there (a corner).

For a piecewise function, first check continuity at the break point; if continuous, compare the slopes of the two pieces there. Example: $x^2$ ($x \le 1$), $2x - 1$ ($x > 1$): both slopes are $2$ at $1$, so differentiable.

Chain rule: if $y = f(u)$ and $u = g(x)$, then $\dfrac{dy}{dx} = \dfrac{dy}{du} \cdot \dfrac{du}{dx}$. Work from the outside in, multiplying one factor per layer: $\dfrac{d}{dx}\sin^3(2x) = 3\sin^2(2x) \cdot \cos(2x) \cdot 2$.

Inverse trigonometric functions (on their principal branches):

• $\dfrac{d}{dx}\sin^{-1} x = \dfrac{1}{\sqrt{1 - x^2}}$, $-1 < x < 1$
• $\dfrac{d}{dx}\cos^{-1} x = -\dfrac{1}{\sqrt{1 - x^2}}$, $-1 < x < 1$
• $\dfrac{d}{dx}\tan^{-1} x = \dfrac{1}{1 + x^2}$, all real $x$

With the chain rule: $\dfrac{d}{dx}\tan^{-1}(2x) = \dfrac{2}{1 + 4x^2}$.

When $x$ and $y$ are related by an equation not solved for $y$, differentiate both sides with respect to $x$, treating $y$ as a function of $x$: $\dfrac{d}{dx}(y^2) = 2y\dfrac{dy}{dx}$, $\dfrac{d}{dx}(xy) = y + x\dfrac{dy}{dx}$, $\dfrac{d}{dx}(\sin y) = \cos y\dfrac{dy}{dx}$. Then collect the $\dfrac{dy}{dx}$ terms and solve.

Example: $x^2 + y^2 = 25$ gives $\dfrac{dy}{dx} = -\dfrac{x}{y}$, which is $-\dfrac34$ at $(3, 4)$. To find a slope at a point, first check the point lies on the curve.

$\dfrac{d}{dx}e^x = e^x$, $\dfrac{d}{dx}\log x = \dfrac1x$ ($x > 0$; $\log$ is the natural logarithm), $\dfrac{d}{dx}a^x = a^x\log a$. With the chain rule: $\dfrac{d}{dx}e^{g(x)} = g'(x)e^{g(x)}$, $\dfrac{d}{dx}\log g(x) = \dfrac{g'(x)}{g(x)}$.

Logarithmic differentiation for $y = u^{v}$ (both depending on $x$) or long products and quotients: take $\log$ of both sides, expand, differentiate (left side $\dfrac{1}{y}\dfrac{dy}{dx}$), then multiply by $y$. Example: $y = x^x$ gives $\dfrac{dy}{dx} = x^x(1 + \log x)$, which is zero at $x = \dfrac1e$. If $y$ is a sum $u + v$, differentiate $u$ and $v$ separately.

Parametric form: if $x = f(t)$ and $y = g(t)$, then $\dfrac{dy}{dx} = \dfrac{dy/dt}{dx/dt}$ (where $\dfrac{dx}{dt} \ne 0$). Example: $x = t^2$, $y = 2t$ gives $\dfrac{dy}{dx} = \dfrac1t$.

Second-order derivative: $\dfrac{d^2y}{dx^2} = \dfrac{d}{dx}\left(\dfrac{dy}{dx}\right)$. For a parametric curve, differentiate $\dfrac{dy}{dx}$ with respect to $t$ and divide by $\dfrac{dx}{dt}$ again.

Relations: to show, say, $y'' - 2y' - 3y = 0$ for $y = e^{3x} + e^{-x}$, find $y'$ and $y''$, substitute and collect like terms. With inverse trigonometric functions, first clear the square root (e.g. $\sqrt{1 - x^2}\,y' = 2y$), then differentiate again.