Continuity and Differentiability — chapter notes

$f$ is continuous at $a$ if $\lim_{x \to a^-} f(x) = \lim_{x \to a^+} f(x) = f(a)$. It is continuous on an interval if it is continuous at every point of it (one-sided at closed ends).

For a piecewise function, only the joins need checking: set the one-sided limits equal to the value and solve for any constants. Standard limits help: $\dfrac{\sin u}{u} \to 1$, $\dfrac{e^u - 1}{u} \to 1$, $\dfrac{1 - \cos u}{u^2} \to \dfrac12$ as $u \to 0$.

If $f$ and $g$ are continuous at $a$, then $f \pm g$, $fg$ and $cf$ are continuous at $a$; $\dfrac{f}{g}$ is continuous at $a$ if $g(a) \ne 0$; and $f \circ g$ is continuous at $a$ if $f$ is continuous at $g(a)$.

So polynomials, $\sin x$, $\cos x$ and $e^x$ are continuous on $\mathbb{R}$; rational functions, $\tan x$, $\sec x$ and $\log x$ are continuous on their domains. If $f$ is continuous at $a$ and $g$ is not, then $f + g$ is not continuous at $a$.

• Removable: $\lim_{x \to a} f(x)$ exists but $f(a)$ is undefined (first type) or $f(a) \ne$ the limit (second type). Remove it by defining $f(a)$ = the limit.
• Jump (non-removable): the one-sided limits are finite but unequal.
• Infinite (non-removable): a one-sided limit is infinite, as for $\dfrac{1}{x - 3}$ at $3$.

Method: factorise and cancel, or use a standard limit, then compare the limits with the value.

Left-hand derivative $Lf'(a) = \lim_{h \to 0^-}\dfrac{f(a + h) - f(a)}{h}$; right-hand derivative $Rf'(a) = \lim_{h \to 0^+}\dfrac{f(a + h) - f(a)}{h}$. $f$ is differentiable at $a$ when both exist and are equal.

Differentiable $\Rightarrow$ continuous, since $f(a + h) - f(a) = \dfrac{f(a + h) - f(a)}{h} \cdot h \to 0$. The converse fails: $|x|$ is continuous at $0$ with one-sided derivatives $-1$ and $1$. $[x]$ is neither continuous nor differentiable at any integer. For a smooth join of two pieces, match both the values and the slopes.

$\dfrac{d}{dx}\sin x = \cos x$, $\dfrac{d}{dx}\cos x = -\sin x$, $\dfrac{d}{dx}\tan x = \sec^2 x$, $\dfrac{d}{dx}\sec x = \sec x\tan x$, $\dfrac{d}{dx}e^x = e^x$, $\dfrac{d}{dx}a^x = a^x\log a$, $\dfrac{d}{dx}\log x = \dfrac{1}{x}$.

Chain rule: if $y = f(u)$ and $u = g(x)$, then $\dfrac{dy}{dx} = f'(u)\,g'(x)$; for three layers multiply three factors. In particular $\dfrac{d}{dx}\log g(x) = \dfrac{g'(x)}{g(x)}$ and $\dfrac{d}{dx}e^{g(x)} = e^{g(x)}g'(x)$.

$\dfrac{d}{dx}\sin^{-1} x = \dfrac{1}{\sqrt{1 - x^2}}$, $\dfrac{d}{dx}\cos^{-1} x = -\dfrac{1}{\sqrt{1 - x^2}}$ ($|x| < 1$); $\dfrac{d}{dx}\tan^{-1} x = \dfrac{1}{1 + x^2}$, $\dfrac{d}{dx}\cot^{-1} x = -\dfrac{1}{1 + x^2}$; $\dfrac{d}{dx}\sec^{-1} x = \dfrac{1}{|x|\sqrt{x^2 - 1}}$, $\dfrac{d}{dx}\operatorname{cosec}^{-1} x = -\dfrac{1}{|x|\sqrt{x^2 - 1}}$ ($|x| > 1$).

Substitution: put $x = \sin\theta$, $\cos\theta$ or $\tan\theta$ to turn the expression into a multiple angle, then use $\sin^{-1}(\sin\phi) = \phi$ only when $\phi$ is in $\left[-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right]$ (and similarly for the others). The interval decides the formula: $\sin^{-1}\dfrac{2x}{1 + x^2} = 2\tan^{-1} x$ for $|x| \le 1$ but $\pi - 2\tan^{-1} x$ for $x > 1$.