The whole Calculus and Optimization formula sheet, to print

Use one-sided limits for $|x|$, floor, piecewise functions and any function that changes formula at $a$.

Only for $x$ in radians and $x\to 0$. For $x\to \infty$, $\sin x/x\to 0$ instead.

Each is a $0/0$ form that equals a derivative at $0$ (of $e^x$, $a^x$ and $\ln (1+x)$ respectively).

Here $n$ is any real number. More generally ${\lim }_{x\to \infty }(1+k/x{)}^x=e^k$ for any real $k$.

Applies when $f(x)>0$ near $a$, $f(x)\to 1$, $g(x)\to \pm \infty$, and the limit inside the exponential exists.

The $e^x$ series holds for all real $x$; the $\ln (1+x)$ series holds for $-1<x\le 1$.

The $\sin x$ and $\cos x$ series hold for all real $x$; the $\tan x$ series for $|x|<\pi /2$. In a $0/0$ limit at $0$, replace each function by enough terms and compare the lowest powers: for example $\frac{x-\sin x}{x^3}=\frac{x^3/6-\cdots }{x^3}\to \frac{1}{6}$.

Needs: $f/g$ of the form $0/0$ or $\pm \infty /\pm \infty$ as $x\to a$; $f,g$ differentiable with $g^{\prime }(x)\ne 0$ on a punctured neighbourhood of $a$; and the limit of $f^{\prime }/g^{\prime }$ (the limit on the right of the equation) to exist, finite or infinite. It also holds for one-sided limits and for $a=\pm \infty$. Differentiate numerator and denominator separately; do not use the quotient rule.

Exponentials beat powers and powers beat logarithms as $x\to \infty$.

Three things must hold: $f(a)$ is defined, the limit exists and is finite, and the two are equal. Equivalently, left limit = right limit = $f(a)$.

Used to show a root exists, for example $f(a)f(b)<0$ gives a root of $f$ in $(a,b)$.

Both conditions matter: on the open interval $(0,1)$, $f(x)=1/x$ is continuous but has no maximum.

$f$ is differentiable at $a$ when this limit exists and is finite.

$f$ is differentiable at $a$ if and only if both exist, are finite and $f_{-}^{\prime }(a)=f_{+}^{\prime }(a)$.

The converse is false. Hence a function that is not continuous at $a$ cannot be differentiable at $a$.

For $u,v$ differentiable at the point.

For $g$ differentiable at $x$ and $f$ differentiable at $g(x)$.

For $u>0$ and $u,v$ differentiable. For example $\frac{d}{dx}{x}^x=x^x(1+\ln x)$ for $x>0$.

All three conditions are needed. Typical use: if $f$ is continuous on the closed interval between two roots and differentiable inside it, then $f^{\prime }$ has a root strictly between them.

Some point has tangent slope equal to the average slope over $[a,b]$. Rolle's theorem is the special case $f(a)=f(b)$.

The condition $g^{\prime }\ne 0$ ensures $g(b)\ne g(a)$ (by Rolle's theorem). Taking $g(x)=x$ gives Lagrange's theorem.

Valid for those $x$ where the series converges to $f(x)$; convergence is not automatic, so state the range of $x$.

The coefficient of $x^n$ is $f^{(n)}(0)/n!$, so $f^{(n)}(0)=n!\times (\text{coefficient of }{x}^n)$.

Converges for all real $x$.

Converges for all real $x$; only odd powers appear.

Converges for all real $x$; only even powers appear.

Valid for $-1<x\le 1$. At $x=1$ it gives $\ln 2$; at $x=-1$ it diverges.

For any real $m$ it is valid for $|x|<1$ and diverges for $|x|>1$ (radius 1) unless $m$ is a non-negative integer, in which case the series stops and is exact for all $x$. At $x=\pm 1$ convergence depends on $m$ (for example, for $m>0$ it also converges at both endpoints).

Valid for $|x|<1$. Replace $x$ by $-x$ for $1/(1+x)=1-x+x^2-\cdots$.

Valid for $-1\le x\le 1$; obtained by integrating $1/(1+t^2)=1-t^2+t^4-\cdots$ from $0$ to $x$. At $x=1$ it gives $\pi /4$.

Valid for $|x|<\pi /2$; only odd powers appear. The first two terms are usually enough for limits as $x\to 0$.

When this limit exists (it may be $\infty$), the series converges absolutely for $|x-a|<R$ and diverges for $|x-a|>R$; the endpoints $|x-a|=R$ must be checked separately. For a series with only odd or only even powers, apply the ratio test to consecutive non-zero terms instead.

Needs $f^{(n)}$ continuous on the closed interval between $a$ and $x$ and $f^{(n+1)}$ to exist on the open interval (having $n+1$ continuous derivatives is enough). Hence $|R_n(x)|\le \frac{M}{(n+1)!}|x-a{|}^{n+1}$, where $M$ is the largest value of $|f^{(n+1)}|$ on that interval.

Applies when, from the first omitted term on, the terms alternate in sign and decrease in magnitude to 0; the error then has the same sign as the first omitted term. Examples: $\ln (1+x)$ for $0<x\le 1$; $\sin x$, $\cos x$ and ${\tan }^{-1}x$ for $|x|\le 1$.

Assume $f^{(n)}$ is continuous near $a$. If $n$ is even: local minimum when $f^{(n)}(a)>0$, local maximum when $f^{(n)}(a)<0$. If $n$ is odd: no local extremum at $a$. For $n=2$ this is the second-derivative test.

Necessary condition: if $f$ has a local extremum at an interior point $c$ and $f^{\prime }(c)$ exists, then $f^{\prime }(c)=0$. So every interior local extremum is a critical point. The converse is false: $f(x)=x^3$ has $f^{\prime }(0)=0$ but no extremum at $0$.

$I$ is an interval, $f$ is continuous on $I$ and differentiable at its interior points. $f^{\prime }\ge 0$ on $I$ gives non-decreasing; if in addition $f^{\prime }=0$ only at isolated points, $f$ is still strictly increasing. The converse of the first implication fails: $x^3$ is strictly increasing on $\mathbb{R}$ although $f^{\prime }(0)=0$.

Needs $f^{\prime }(c)=0$ and $f^{\prime \prime }(c)$ to exist, so it cannot be used at a critical point where $f^{\prime }(c)$ does not exist. If $f^{\prime \prime }(c)=0$ the test is inconclusive and you must use the first-derivative test or higher derivatives.

Both conditions matter: continuity, and a closed bounded interval. On an open interval an extremum may not exist, for example $f(x)=x$ on $(0,1)$ has neither. A discontinuous function on a closed interval may also fail, for example $f(x)=x$ for $0\le x<1$ with $f(1)=0$ has no maximum on $[0,1]$.

Convex (also called concave up) means $f(\lambda x+(1-\lambda )y)\le \lambda f(x)+(1-\lambda )f(y)$ for all $x,y\in I$ and $\lambda \in [0,1]$. For differentiable $f$ this is equivalent to the graph lying on or above every tangent line: $f(y)\ge f(x)+f^{\prime }(x)(y-x)$ for all $x,y\in I$. $f^{\prime \prime }>0$ on $I$ makes $f$ strictly convex, but not conversely: $x^4$ is strictly convex although $f^{\prime \prime }(0)=0$.

A point of inflection is a point $c$ where $f$ is continuous and the graph changes from convex to concave or the reverse. If $f^{\prime \prime }(c)$ exists at an inflection point then $f^{\prime \prime }(c)=0$. The converse is false: $f(x)=x^4$ has $f^{\prime \prime }(0)=0$ but $f^{\prime \prime }\ge 0$ on both sides, so $0$ is not an inflection point. An inflection point can also occur where $f^{\prime \prime }$ does not exist: $f(x)=x|x|$ has $f^{\prime \prime }(x)=-2$ for $x<0$ and $2$ for $x>0$, with $f^{\prime \prime }(0)$ undefined.

The same holds with min in place of max. Reason: if some $x_1\in I$ had $f(x_1)>f(c)$, $f$ would have a local minimum strictly between $c$ and $x_1$, which would be a second critical point.

The iterate stays put exactly when $f^{\prime }(x_k)=0$, that is at a critical point. That point need not be a minimum if the iterate lands on it exactly. On a non-convex $f$ the iteration can settle at a local minimum that is not global; which one depends on $x_0$ and $\eta$.

Converges to $x^{\ast }$ from every starting point if and only if $|1-\eta a|<1$, that is $0<\eta <2/a$. With $\eta =1/a$ it reaches $x^{\ast }$ in one step. For $1/a<\eta <2/a$ the iterates alternate sides of $x^{\ast }$ while converging. At $\eta =2/a$ they bounce between $x_0$ and $2x^{\ast }-x_0$ forever, and for $\eta >2/a$ they diverge (whenever $x_0\ne x^{\ast }$). Here $a=f^{\prime \prime }$, the constant curvature.