Course Review

$\displaystyle\lim_{x\to a}f(x)=L$ means $f(x)$ can be made arbitrarily close to $L$ by taking $x$ close enough to $a$ (but $x\neq a$). The value $f(a)$ is irrelevant to the limit --- only the behavior near $a$ matters.

If $\displaystyle\lim_{x\to a}f=L$ and $\displaystyle\lim_{x\to a}g=M$ (both finite), then multicols2

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• $\lim(f\pm g)=L\pm M$
• $\lim(cf)=cL$
• $\lim(fg)=LM$
• $\lim\dfrac{f}{g}=\dfrac{L}{M}$ (if $M\neq0$)
• $\lim f^{\,n}=L^{n}$
• $\lim\sqrt[n]{f}=\sqrt[n]{L}$ ($L\ge0$ if $n$ even)

multicols For continuous functions, $\displaystyle\lim_{x\to a}f(x)=f(a)$ (direct substitution).

One-sided: $\displaystyle\lim_{x\to a^-}f$ (from the left) and $\displaystyle\lim_{x\to a^+}f$ (from the right). The two-sided limit exists $\iff$ both one-sided limits exist and are equal. Infinite limits ($\to\pm\infty$) signal a vertical asymptote at $x=a$. Limits at infinity describe end behavior and horizontal asymptotes: for a rational function, compare degrees --- if $\deg(\text{num})<\deg(\text{den})$ the limit is $0$; if equal, it is the ratio of leading coefficients; if greater, it is $\pm\infty$ (no horizontal asymptote).

Direct substitution sometimes yields an indeterminate form --- no immediate answer, more work required:

Resolve with algebra (factor/cancel, rationalize, common denominators), the Squeeze Theorem, or L'Hôpital's Rule. Forms like $\frac{k}{0}$ ($k\neq0$) are not indeterminate --- they give $\pm\infty$.

$f$ is continuous at $a$ if all three hold: (1) $f(a)$ is defined, (2) $\displaystyle\lim_{x\to a}f(x)$ exists, and (3) $\displaystyle\lim_{x\to a}f(x)=f(a)$. IVT: If $f$ is continuous on $[a,b]$ and $N$ is any value between $f(a)$ and $f(b)$, then there exists $c\in(a,b)$ with $f(c)=N$. (Guarantees a root when $f(a),f(b)$ have opposite signs.)

Squeeze: If $g(x)\le f(x)\le h(x)$ near $a$ and $\displaystyle\lim_{x\to a}g=\lim_{x\to a}h=L$, then $\displaystyle\lim_{x\to a}f=L$.

The derivative is the instantaneous rate of change / slope of the tangent line:

Notations: $f'(x)$, $\dfrac{dy}{dx}$, $y'$, $D_x f$, $\dfrac{d}{dx}f$.

Differentiable $\Rightarrow$ continuous, but not conversely. A function fails to be differentiable at corners (e.g. $|x|$ at $0$), cusps, vertical tangents (infinite slope), and any discontinuity.

multicols2

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• Constant: $\dfrac{d}{dx}[c]=0$
• Power: $\dfrac{d}{dx}[x^{n}]=n x^{n-1}$
• Constant multiple: $(cf)'=cf'$
• Sum: $(f\pm g)'=f'\pm g'$
• Product: $(fg)'=f'g+fg'$
• Quotient: $\left(\dfrac{f}{g}\right)'=\dfrac{f'g-fg'}{g^{2}}$

multicols

multicols2

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• $\dfrac{d}{dx}e^{x}=e^{x}$
• $\dfrac{d}{dx}a^{x}=a^{x}\ln a$
• $\dfrac{d}{dx}\ln x=\dfrac{1}{x}$
• $\dfrac{d}{dx}\log_a x=\dfrac{1}{x\ln a}$
• $\dfrac{d}{dx}\sin x=\cos x$
• $\dfrac{d}{dx}\cos x=-\sin x$
• $\dfrac{d}{dx}\tan x=\sec^2 x$
• $\dfrac{d}{dx}\cot x=-\csc^2 x$
• $\dfrac{d}{dx}\sec x=\sec x\tan x$
• $\dfrac{d}{dx}\csc x=-\csc x\cot x$

multicols

The derivative of a composition $f(g(x))$ is

In Leibniz form: $\dfrac{dy}{dx}=\dfrac{dy}{du}\cdot\dfrac{du}{dx}$. General power rule: $\dfrac{d}{dx}[u^{n}]=n u^{n-1}u'$.

When $y$ is defined implicitly by an equation, differentiate both sides with respect to $x$, applying the chain rule to every $y$ term (each contributes a factor $\frac{dy}{dx}$), then solve algebraically for $\frac{dy}{dx}$.

If $f$ is one-to-one and differentiable with $f(a)=b$ (so $f^{-1}(b)=a$), then

The slope of the inverse is the reciprocal of the slope of $f$ at the corresponding point.

multicols2

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• $\dfrac{d}{dx}\arcsin x=\dfrac{1}{\sqrt{1-x^2}}$
• $\dfrac{d}{dx}\arccos x=-\dfrac{1}{\sqrt{1-x^2}}$
• $\dfrac{d}{dx}\arctan x=\dfrac{1}{1+x^2}$
• $\dfrac{d}{dx}\operatorname{arccot} x=-\dfrac{1}{1+x^2}$
• $\dfrac{d}{dx}\operatorname{arcsec} x=\dfrac{1}{|x|\sqrt{x^2-1}}$
• $\dfrac{d}{dx}\operatorname{arccsc} x=-\dfrac{1}{|x|\sqrt{x^2-1}}$

multicols

For products of many factors, or a variable base with a variable exponent $y=[f(x)]^{g(x)}$: take $\ln$ of both sides, use log laws to expand, differentiate implicitly, then multiply back by $y$.

A derivative $\frac{dy}{dx}$ carries units of “$y$-units per $x$-unit.” In context, $f'(a)$ is the instantaneous rate of change of $f$ at $a$; interpret answers with correct units (e.g. “the volume is increasing at $12$ cm$^3$/sec”).

For position $s(t)$: velocity $v(t)=s'(t)$; acceleration $a(t)=v'(t)=s''(t)$; speed $=|v(t)|$.

• [leftmargin=*,topsep=1pt,itemsep=1pt]
• Particle moves right/up when $v>0$, left/down when $v<0$, at rest when $v=0$.
• Speeding up when $v$ and $a$ have the same sign; slowing down when opposite signs.
• Displacement $=\int_a^b v\,dt$; total distance $=\int_a^b |v|\,dt$.

When several quantities change with time and are linked by an equation: differentiate the relating equation implicitly with respect to $t$, then substitute known values and solve for the unknown rate. Substitute numbers only after differentiating.

The tangent-line (linear) approximation near $x=a$:

Differentials: $dy=f'(x)\,dx$ estimates the change $\Delta y$ for a small change $dx=\Delta x$. Overestimates where $f$ is concave down, underestimates where concave up.

If $\displaystyle\lim\frac{f}{g}$ is of form $\frac00$ or $\frac{\infty}{\infty}$ (and $g'\neq0$ nearby), then

provided the right side exists. For $0\cdot\infty$ or $\infty-\infty$, rewrite as a quotient first; for $1^{\infty},0^0,\infty^0$, take $\ln$ first.

MVT: If $f$ is continuous on $[a,b]$ and differentiable on $(a,b)$, then there is $c\in(a,b)$ with

Rolle's Theorem is the special case $f(a)=f(b)$: then $f'(c)=0$ for some $c$.

$f$ increases where $f'>0$, decreases where $f'<0$. At a critical point ($f'=0$ or undefined), $f$ has a

• [leftmargin=*,topsep=1pt,itemsep=1pt]
• local max if $f'$ changes $+\to-$,
• local min if $f'$ changes $-\to+$,
• neither if $f'$ does not change sign.

$f$ is concave up where $f''>0$, concave down where $f''<0$. An inflection point is where concavity changes (requires $f''=0$ or undefined and a sign change). Second Derivative Test: at a critical point $c$ with $f'(c)=0$: if $f''(c)>0$ it's a local min; if $f''(c)<0$ a local max; if $f''(c)=0$ the test is inconclusive.

EVT: a function continuous on a closed interval $[a,b]$ attains an absolute max and min there. Find them with the Candidates Test: evaluate $f$ at all critical points and at the endpoints $a,b$; the largest/smallest outputs are the absolute extrema.

Optimization: write the quantity to optimize as a function of one variable (use a constraint to eliminate others), find critical points, and justify the extremum. Connections: $f'>0\Rightarrow f\uparrow$; $f''>0\Rightarrow f$ concave up. A max of $f'$ occurs where $f''=0$ changes $+\to-$ (an inflection point of $f$). Reading any one of $f,f',f''$ constrains the others.

multicols2

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• $\displaystyle\int x^{n}\,dx=\frac{x^{n+1}}{n+1}+C$ ($n\neq-1$)
• $\displaystyle\int \frac{1}{x}\,dx=\ln|x|+C$
• $\displaystyle\int e^{x}\,dx=e^{x}+C$
• $\displaystyle\int a^{x}\,dx=\frac{a^{x}}{\ln a}+C$
• $\displaystyle\int \sin x\,dx=-\cos x+C$
• $\displaystyle\int \cos x\,dx=\sin x+C$
• $\displaystyle\int \sec^2 x\,dx=\tan x+C$
• $\displaystyle\int \sec x\tan x\,dx=\sec x+C$
• $\displaystyle\int \frac{dx}{\sqrt{1-x^2}}=\arcsin x+C$
• $\displaystyle\int \frac{dx}{1+x^2}=\arctan x+C$

multicols

A definite integral is a limit of Riemann sums: $\displaystyle\int_a^b f\,dx=\lim_{n\to\infty}\sum_{i=1}^{n}f(x_i^*)\,\Delta x$, $\Delta x=\frac{b-a}{n}$. Approximations (left/right/midpoint) sample $x_i^*$ at those points. Trapezoidal Rule (equal width $\Delta x$):

Trapezoid overestimates when concave up, underestimates when concave down.

multicols2

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• $\displaystyle\int_a^a f\,dx=0$
• $\displaystyle\int_b^a f\,dx=-\int_a^b f\,dx$
• $\displaystyle\int_a^b cf\,dx=c\int_a^b f\,dx$
• $\displaystyle\int_a^b (f\pm g)\,dx=\int_a^b f\,dx\pm\int_a^b g\,dx$
• $\displaystyle\int_a^b f\,dx+\int_b^c f\,dx=\int_a^c f\,dx$

multicols

Part 1 (accumulation / derivative of an integral): if $\displaystyle g(x)=\int_a^{x} f(t)\,dt$ with $f$ continuous, then $g'(x)=f(x)$. With a variable upper limit $u(x)$ (chain rule): $\dfrac{d}{dx}\displaystyle\int_a^{u(x)}f(t)\,dt=f\big(u(x)\big)\,u'(x)$. Part 2 (evaluation): if $F'=f$, then $\displaystyle\int_a^b f(x)\,dx=F(b)-F(a)$.

$u$-sub reverses the chain rule: let $u=g(x)$, $du=g'(x)\,dx$, and change the limits for a definite integral. Average value of $f$ on $[a,b]$: $\displaystyle f_{\text{avg}}=\frac{1}{b-a}\int_a^b f(x)\,dx$. Net Change Theorem: $\displaystyle\int_a^b F'(x)\,dx=F(b)-F(a)$ --- the integral of a rate gives the total (net) accumulated change.

Choose $u$ by LIATE priority --- Logarithmic, Inverse trig, Algebraic, Trigonometric, Exponential --- whichever comes first is $u$; the rest is $dv$. Repeat for higher powers; use tabular integration to organize.

For a proper rational function (numerator degree $<$ denominator), factor the denominator and split into simpler fractions:

solve for constants, then integrate each piece (giving logs / arctangents). If improper, do polynomial long division first.

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• $\int\sin^m x\cos^n x\,dx$: if a power is odd, peel one factor and convert with $\sin^2+\cos^2=1$.
• If both powers even, use $\sin^2 x=\frac{1-\cos2x}{2}$, $\cos^2 x=\frac{1+\cos2x}{2}$.
• $\int\tan^m x\sec^n x\,dx$: use $\sec^2=1+\tan^2$ with $\frac{d}{dx}\tan x=\sec^2 x$.

multicols3

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• $\sqrt{a^2-x^2}$: let $x=a\sin\theta$
• $\sqrt{a^2+x^2}$: let $x=a\tan\theta$
• $\sqrt{x^2-a^2}$: let $x=a\sec\theta$

multicols Then simplify the radical with a Pythagorean identity and convert back with a reference triangle.

Handle infinite limits or infinite discontinuities as limits: $\displaystyle\int_a^{\infty}f\,dx=\lim_{b\to\infty}\int_a^{b}f\,dx$. Converges if the limit is finite, else diverges. Key result:

Between $y=f(x)$ (top) and $y=g(x)$ (bottom) on $[a,b]$:

If curves are given as $x=$ functions of $y$, integrate (right $-$ left) $dy$. Find intersection points to set the limits.

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• Disks: $\displaystyle V=\pi\int_a^b [R(x)]^2\,dx$ (solid region against the axis).
• Washers: $\displaystyle V=\pi\int_a^b \big([R]^2-[r]^2\big)\,dx$ (outer $R$, inner $r$; gap from the axis).
• Shells: $\displaystyle V=2\pi\int_a^b (\text{radius})(\text{height})\,dx$ (convenient when the axis is parallel to the slices).

If cross-sections perpendicular to the $x$-axis have area $A(x)$:

Squares: $A=s^2$; equilateral triangles: $A=\frac{\sqrt3}{4}s^2$; semicircles: $A=\frac{\pi}{8}s^2$ (with $s=$ base length from the region).

Arc length of $y=f(x)$ on $[a,b]$: $\displaystyle L=\int_a^b\sqrt{1+[f'(x)]^2}\,dx$. Average value: $\displaystyle f_{\text{avg}}=\frac{1}{b-a}\int_a^b f\,dx$. Motion: displacement $=\int v\,dt$; total distance $=\int|v|\,dt$; $s(t)=s(t_0)+\int_{t_0}^{t}v\,d\tau$.

A slope field plots short segments with slope $\frac{dy}{dx}$ at grid points, showing the family of solution curves. A particular solution follows the segments through its initial point. Slopes depend only on where $\frac{dy}{dx}$ is evaluated.

If $\dfrac{dy}{dx}=g(x)h(y)$, separate and integrate:

Add $+C$ once, then apply the initial condition to solve for $C$ and get the particular solution.

$\dfrac{dP}{dt}=kP$ (rate proportional to amount) solves to

with $k>0$ growth, $k<0$ decay, $P_0=P(0)$. Half-life / doubling problems reduce to solving for $t$.

Growth limited by a carrying capacity $M$:

Key facts: $\displaystyle\lim_{t\to\infty}P=M$; growth is fastest at $P=\frac{M}{2}$ (the inflection point of $P(t)$); solutions are $S$-shaped. Solution: $P(t)=\dfrac{M}{1+A e^{-kt}}$ with $A=\dfrac{M-P_0}{P_0}$.

Numerically approximate a solution from a starting point using step size $h$:

where $f(x,y)=\frac{dy}{dx}$. Follow the tangent line one step at a time.

Geometric: $\displaystyle\sum_{n=0}^{\infty}ar^{n}$ converges $\iff |r|<1$, with sum $\dfrac{a}{1-r}$ ($a=$ first term). $p$-series: $\displaystyle\sum_{n=1}^{\infty}\frac{1}{n^{p}}$ converges $\iff p>1$. The harmonic series ($p=1$) diverges.

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• $n$th-Term Test: if $\lim a_n\neq0$, diverges. (Can only show divergence.)
• Integral Test: $f$ positive, decreasing, continuous $\Rightarrow \sum a_n$ and $\int_1^{\infty}f\,dx$ share fate.
• Direct Comparison: $0\le a_n\le b_n$; bigger converges $\Rightarrow$ smaller converges (and dually).
• Limit Comparison: $\lim\frac{a_n}{b_n}=L$, $0<L<\infty\Rightarrow$ same behavior as $\sum b_n$.
• Ratio Test: $L=\lim\big|\frac{a_{n+1}}{a_n}\big|$; $L<1$ converges, $L>1$ diverges, $L=1$ inconclusive.
• Root Test: $L=\lim\sqrt[n]{|a_n|}$; same conclusions as ratio test.
• Alternating Series Test: $\sum(-1)^n b_n$ with $b_n\downarrow0$ converges.

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• $n$th-term: does $a_n\to0$? If not $\Rightarrow$ diverges. (Always first.)
• Geometric ($ar^n$)? converges iff $|r|<1$, sum $\frac{a}{1-r}$.
• $p$-series ($\frac{1}{n^p}$)? converges iff $p>1$.
• Factorials or $n$th powers? Ratio (or Root) Test.
• Rational / resembles a $p$-series? Direct or Limit Comparison.
• Positive, decreasing, easy to integrate? Integral Test.
• Alternating $(-1)^n b_n$? AST; then check $\sum|a_n|$ for absolute vs. conditional.

Alternating Series Error: if AST holds, $|S-S_N|\le b_{N+1}$ (first omitted term). Absolute convergence: $\sum|a_n|$ converges (implies $\sum a_n$ converges). Conditional: $\sum a_n$ converges but $\sum|a_n|$ diverges.

$\displaystyle\sum a_n(x-c)^n$ converges on $|x-c|<R$. Find $R$ by the ratio test; then test each endpoint separately (ratio test gives $L=1$ there). Term-by-term differentiation/integration keeps the same $R$ (endpoints may change).

Taylor about $c$: $\displaystyle f(x)=\sum_{n=0}^{\infty}\frac{f^{(n)}(c)}{n!}(x-c)^n$. Maclaurin is $c=0$. Lagrange Error Bound: $\displaystyle |R_n(x)|\le\frac{M}{(n+1)!}\,|x-c|^{\,n+1}$, where $M$ bounds $|f^{(n+1)}|$ between $c$ and $x$.