GRE Mathematics Subject Test

Limits

The statement $\lim_{x \to a} f(x) = L$ means that $f(x)$ can be made as close to $L$ as we like by taking $x$ close enough to $a$, without $x$ being equal to $a$. The function need not be defined at $a$, and even if it is, its value there plays no part in the limit. In the formal version, for every $\varepsilon > 0$ there is a $\delta > 0$ such that $0 < |x - a| < \delta$ implies $|f(x) - L| < \varepsilon$.

A one-sided limit approaches $a$ from one side only, written $\lim_{x \to a^-}$ and $\lim_{x \to a^+}$. The two-sided limit exists exactly when both one-sided limits exist and are equal. For $f(x) = |x|/x$ the left limit at $0$ is $-1$ and the right limit is $1$, so there is no limit at $0$.

A limit at infinity describes behaviour as $x$ grows without bound, and an infinite limit means the values grow without bound. For a rational function, divide numerator and denominator by the highest power of $x$ in the denominator. For example

Limit laws

If $\lim_{x \to a} f(x) = L$ and $\lim_{x \to a} g(x) = M$ both exist, then the limit of a sum, difference, product and constant multiple is the corresponding combination of $L$ and $M$, the limit of a quotient is $L/M$ provided $M \neq 0$, and $\lim f(x)^n = L^n$. If $f$ is continuous at $M$ then $\lim f(g(x)) = f(M)$.

The squeeze theorem

If $g(x) \le f(x) \le h(x)$ near $a$ (except possibly at $a$) and $\lim_{x\to a} g(x) = \lim_{x\to a} h(x) = L$, then $\lim_{x \to a} f(x) = L$.

Because $-1 \le \sin(1/x) \le 1$, multiplying by $x^2$ gives $-x^2 \le x^2 \sin(1/x) \le x^2$, and both bounds tend to $0$, so $\lim_{x \to 0} x^2\sin(1/x) = 0$. The same function with $x$ in place of $x^2$ is squeezed in the same way, but $\sin(1/x)$ alone has no limit at $0$ because it oscillates between $-1$ and $1$ infinitely often.

Standard limits

With $x$ in radians,

Exponentials beat powers and powers beat logarithms: for any $k > 0$, $x^k / e^x \to 0$ and $(\ln x)/x^k \to 0$ as $x \to \infty$.

Indeterminate forms and L'Hôpital's rule

The forms $\frac{0}{0}$, $\frac{\infty}{\infty}$, $0 \cdot \infty$, $\infty - \infty$, $1^\infty$, $0^0$ and $\infty^0$ do not determine a limit by themselves. L'Hôpital's rule handles the first two: if $f$ and $g$ are differentiable near $a$, $\lim f = \lim g = 0$ (or both tend to $\pm\infty$), and $\lim \frac{f'(x)}{g'(x)}$ exists or is infinite, then

It can be applied repeatedly. For example

The other forms are first rewritten as a quotient. For $0\cdot\infty$, write $x \ln x = \frac{\ln x}{1/x}$ as $x \to 0^+$; this is of the form $\infty/\infty$, and the rule gives $\frac{1/x}{-1/x^2} = -x \to 0$. For the forms $1^\infty$, $0^0$ and $\infty^0$, take logarithms: if $y = f(x)^{g(x)}$ then $\ln y = g(x)\ln f(x)$, find the limit of that, and exponentiate. Thus for $y = x^x$ as $x \to 0^+$, $\ln y = x\ln x \to 0$, so $y \to e^0 = 1$.

Continuity

A function $f$ is continuous at $a$ if $f(a)$ is defined, $\lim_{x \to a} f(x)$ exists, and the two are equal. It is continuous on an interval if it is continuous at every point there (at an endpoint, using the one-sided limit). Polynomials, $e^x$, $\sin x$ and $\cos x$ are continuous everywhere, rational functions are continuous wherever the denominator is non-zero, and sums, products, quotients (with non-zero denominator) and compositions of continuous functions are continuous.

There are three ways for continuity to fail at $a$.

A removable discontinuity is repaired by redefining $f(a)$ as the limit.

Theorems for continuous functions on a closed interval

Let $f$ be continuous on $[a, b]$.

• The intermediate value theorem: $f$ takes every value between $f(a)$ and $f(b)$ at some point of $[a,b]$. In particular, if $f(a)$ and $f(b)$ have opposite signs, $f$ has a zero in $(a, b)$.
• The extreme value theorem: $f$ attains a maximum and a minimum on $[a,b]$. Both hypotheses are needed: $f(x) = 1/x$ on $(0, 1]$ has no maximum, and $f(x) = x$ on $(0,1)$ has no maximum or minimum.
• $f$ is bounded on $[a, b]$, and is in fact uniformly continuous there.

As an example of the intermediate value theorem, $f(x) = x^3 + x - 1$ has $f(0) = -1 < 0$ and $f(1) = 1 > 0$, so it has a root in $(0, 1)$. A continuous map from $[0,1]$ to itself has a fixed point, since $g(x) = f(x) - x$ satisfies $g(0) \ge 0$ and $g(1) \le 0$.

Continuity and differentiability

A function differentiable at $a$ is continuous at $a$, because $f(a + h) - f(a) = h \cdot \frac{f(a+h) - f(a)}{h} \to 0 \cdot f'(a) = 0$. The converse fails: $|x|$ is continuous at $0$ but has a corner there, with one-sided gradients $-1$ and $1$.

Limits of sequences

A sequence $a_n \to L$ if $|a_n - L|$ eventually stays below any given positive number. A monotonic bounded sequence converges. If $f$ is continuous and $a_{n+1} = f(a_n)$ with $a_n \to L$, then $L = f(L)$. For $a_1 = 1$, $a_{n+1} = \sqrt{2 + a_n}$, the terms increase and are bounded above by $2$ (if $a_n < 2$ then $a_{n+1} < \sqrt 4 = 2$), so they converge to a limit satisfying $L = \sqrt{2 + L}$, that is $L^2 - L - 2 = 0$, and since $L > 0$, $L = 2$.