Limits and infinite limits

The limit is the basic concept of calculus. The derivative is a limit, the integral is a limit, and continuity is defined with a limit. On exams the limit questions are usually on the first page: quick points. But each one hides a small trap. This week’s algorithm is simple:

1. Substitute first. If you get a finite number, you are done.
2. If you get $\tfrac00$, simplify with algebra: factor, multiply by the conjugate, or build the $\tfrac{\sin x}{x}$ pattern.
3. If you get $\tfrac{c}{0}$ ($c\ne0$), the answer is $\pm\infty$ or “does not exist”. Check the sign from each side.
4. If there is an absolute value, a piecewise function or a root, compute the right and left limits separately.
5. If there is an oscillating expression (like $\sin\tfrac1x$), use the Squeeze Theorem.

The meaning of a limit

$\lim_{x\to a}f(x)=L$

means “when $x$ gets close enough to $a$, $f(x)$ gets as close to $L$ as you want”. The critical point: the value of $f(a)$ does not affect the limit. $f(a)$ may be undefined, or it may be some other number. The limit only looks around $a$.

For example, $f(x)=\dfrac{x^2-1}{x-1}$ is undefined at $x=1$. But for $x=0.999$ it gives $1.999$, and for $x=1.001$ it gives $2.001$. The limit is $2$.

One-sided limits

• $\displaystyle\lim_{x\to a^-}f(x)$: as $x$ approaches $a$ from the left ($x<a$).
• $\displaystyle\lim_{x\to a^+}f(x)$: from the right ($x>a$).

$\lim_{x\to a}f(x)=L \iff \lim_{x\to a^-}f(x)=\lim_{x\to a^+}f(x)=L$

Example 1. The function $\dfrac{|x|}{x}$ is $1$ for $x>0$ and $-1$ for $x<0$. The right limit is $1$ and the left limit is $-1$. $\lim_{x\to0}$ does not exist.

Example 1b (piecewise function). $\lim_{x\to1}f(x)$ for $f(x)=\begin{cases}x^2+1, & x<1\\ 3x-1, & x\ge1\end{cases}$.

From the left the piece $x<1$ is used: $1^2+1=2$. From the right the piece $x\ge1$: $3-1=2$. Both sides are equal, so the limit is $2$. At a junction point always work like this: for each direction, use the formula of that direction.

Three cases where the limit does not exist

1. The left and right limits differ (a jump): $\dfrac{|x|}{x}$, $x\to0$.
2. Unbounded growth: $\dfrac1x$, $x\to0$. $+\infty$ on one side, $-\infty$ on the other.
3. Oscillation: $\sin\dfrac1x$, $x\to0$. At the points $x=\tfrac{1}{n\pi}$ the value is $0$, and at the points $x=\tfrac{2}{(4n+1)\pi}$ it is $1$. As it approaches $0$, the function swings between $-1$ and $1$ infinitely many times and never settles on a single number.

Guessing a limit with a numerical table is misleading in this third case. You see different numbers depending on which points you choose. A table gives an idea, not a proof.

Limit laws

If $\lim f=L$ and $\lim g=M$: the limit of a sum is $L+M$, of a product $LM$, of a quotient $L/M$ (if $M\ne0$), $\lim[f]^n=L^n$, $\lim\sqrt[n]{f}=\sqrt[n]{L}$. Consequence: for polynomials and rational functions (as long as the denominator is not zero), substitution is enough.

Example 2. $\displaystyle\lim_{x\to2}(x^2+3x)=4+6=10$.

All the laws work under the condition “if the limits exist”. If one of the two limits does not exist, you cannot use the sum law; this also does not mean that the limit of the sum does not exist. For example $\lim_{x\to0}\sin\tfrac1x$ does not exist and $\lim_{x\to0}\left(-\sin\tfrac1x\right)$ does not exist, but their sum is $0$ everywhere and its limit is $0$.

The typical flow on an exam is:

substitute
 ├─ got a number          -> answer
 ├─ c/0 (c ≠ 0)           -> ±inf or does not exist: left/right sign analysis
 └─ 0/0                   -> algebra:
      ├─ polynomial/rational -> factor
      ├─ difference of roots -> multiply by the conjugate
      ├─ sin, tan, cos       -> sin(u)/u pattern
      └─ oscillation × 0     -> squeeze

Resolving 0/0

$\tfrac00$ is not an answer, it is a sign that says “work more” (an indeterminate form). There are three main techniques.

1. Factoring

Example 3. $\displaystyle\lim_{x\to-1}\frac{x^2+4x+3}{x+1}=\lim_{x\to-1}\frac{(x+1)(x+3)}{x+1}=\lim_{x\to-1}(x+3)=2$.

If the denominator is zero as $x\to a$ and the numerator is also zero, both have a factor $(x-a)$. Find it and cancel it.

2. Multiplying by the conjugate (rationalizing)

For differences that contain roots, multiply the expression $\sqrt{A}-B$ by $\sqrt{A}+B$ to get $(\sqrt A)^2-B^2$.

Example 4.

$\lim_{x\to0}\frac{\sqrt{x+9}-3}{x}=\lim_{x\to0}\frac{(x+9)-9}{x(\sqrt{x+9}+3)}=\lim_{x\to0}\frac{1}{\sqrt{x+9}+3}=\frac16$

3. The trigonometric pattern

The basic limit in radians:

$\lim_{x\to0}\frac{\sin x}{x}=1$

Derived from it: $\displaystyle\lim_{x\to0}\frac{\tan x}{x}=1$ and $\displaystyle\lim_{x\to0}\frac{1-\cos x}{x}=0$.

Example 5. $\displaystyle\lim_{x\to0}\frac{\sin 4x}{x}=\lim_{x\to0}4\cdot\frac{\sin 4x}{4x}=4$.

Rule: the expression inside the sine must also appear exactly in the denominator. If it does not, create it by multiplying and dividing. Short result: $\dfrac{\sin ax}{bx}\to\dfrac{a}{b}$.

Example 5b. $\displaystyle\lim_{x\to0}\frac{\sin 3x}{\sin 5x}$. Put the numerator and denominator into the pattern separately:

$\frac{\sin 3x}{\sin 5x}=\frac{\dfrac{\sin 3x}{3x}\cdot 3x}{\dfrac{\sin 5x}{5x}\cdot 5x}=\frac{3}{5}\cdot\frac{\dfrac{\sin 3x}{3x}}{\dfrac{\sin 5x}{5x}}\ \to\ \frac35\cdot\frac11=\frac35$

For expressions with $\tan$, write $\tan x=\dfrac{\sin x}{\cos x}$; since $\cos x\to1$, $\dfrac{\tan ax}{bx}\to\dfrac ab$ also holds.

PITFALL · the pattern does not work unless $x\to0$. $\displaystyle\lim_{x\to\pi}\frac{\sin x}{x}=\frac{0}{\pi}=0$, not $1$. The pattern holds only when the inside of the sine goes to zero.

The Squeeze Theorem

If $g(x)\le f(x)\le h(x)$ near $a$ and $\lim g=\lim h=L$, then $\lim f=L$.

The most common use: a bounded expression times an expression that goes to zero.

Example 6. $\displaystyle\lim_{x\to0}x^2\cos\frac1x$.

$\cos\tfrac1x$ has no limit (it oscillates constantly). But $-1\le\cos\tfrac1x\le1$. Multiply by $x^2\ge0$:

$-x^2\le x^2\cos\frac1x\le x^2$

Both ends go to $0$, so the middle one also goes to $0$.

PITFALL · using a limit law in the wrong place. You cannot write "$\lim x^2\cdot\lim\cos\tfrac1x=0\cdot(\text{?})$"; the product law requires both limits to exist. Use the squeeze.

Infinite limits and vertical asymptotes

If $f(x)$ grows without bound as $x\to a$, we write $\lim f=+\infty$. This does not mean the limit “exists”; it describes the behavior. The line $x=a$ is a vertical asymptote.

In the case $\tfrac{c}{0}$ ($c\ne0$), look at the sign of the denominator to determine the sign:

Example 7. $\displaystyle\lim_{x\to2^-}\frac{x+1}{x-2}$. The numerator $\to3>0$. For $x<2$, $x-2<0$. The result is $\tfrac{3}{0^-}=-\infty$. From the right it is $+\infty$.

Example 8. $\displaystyle\lim_{x\to3}\frac{1}{(x-3)^2}=+\infty$. The square is positive on both sides, so the two-sided limit is $+\infty$.

Algorithm for finding vertical asymptotes (rational function):

1. Find the x values that make the denominator zero.
2. Factor the numerator and denominator, and cancel common factors.
3. A point that cancels -> a hole (removable), NOT an asymptote.
4. A point that still makes the denominator zero after cancelling -> vertical asymptote.

The precise definition of a limit (epsilon-delta)

$\displaystyle\lim_{x\to a}f(x)=L$ means: for every $\varepsilon>0$ there is a $\delta>0$ such that $0<|x-a|<\delta$ implies $|f(x)-L|<\varepsilon$.

$\varepsilon$ is “how much error I accept in the output”, and $\delta$ is “how close I must keep the input”. For linear functions the computation is short:

Example 9. $\displaystyle\lim_{x\to1}(3x+2)=5$. $|(3x+2)-5|=3|x-1|$. For this to be smaller than $\varepsilon$, it is enough that $|x-1|<\varepsilon/3$. Choose $\delta=\varepsilon/3$.

In Adams this definition comes with numerical questions of the type “find $\delta$ for $\varepsilon=0.01$”. Rearrange the difference so that it contains a factor $|x-a|$, and divide by the coefficient.

Common mistakes