AP Calculus BC

What a limit says

The limit of $f(x)$ as $x$ approaches $c$ is the number $R$ that $f(x)$ can be made arbitrarily close to by taking $x$ sufficiently close to $c$, without letting $x$ equal $c$. It is written

The value of $f$ at $c$ plays no part. The function may be undefined at $c$, or defined with a different value, and the limit is still decided only by what happens nearby. That is why limits are the tool for describing the behaviour of $\frac{x^2 - 9}{x - 3}$ at $x = 3$, where the expression itself is not defined.

A limit is a real number or it does not exist. Writing $\lim_{x \to c} f(x) = \infty$ describes the way in which a limit fails to exist, and it is not a value.

One-sided limits

$\lim_{x \to c^-} f(x)$ is the limit as $x$ approaches $c$ from the left, using only values below $c$. $\lim_{x \to c^+} f(x)$ uses only values above $c$. The two-sided limit exists exactly when both one-sided limits exist and are equal:

For $f(x) = 2x - 1$ when $x < 3$ and $f(x) = x^2 - 6$ when $x \ge 3$, the left limit at $3$ is $2(3) - 1 = 5$ and the right limit is $9 - 6 = 3$. They differ, so $\lim_{x \to 3} f(x)$ does not exist, even though $f(3) = 3$ is defined.

Ways a limit can fail to exist

There are three common reasons.

• The one-sided limits exist but are different, as in the jump above. A graph shows this as a break with the two sides ending at different heights.
• The function is unbounded near $c$. For $\frac{1}{x}$ at $0$ the values grow without bound on the right and fall without bound on the left.
• The function oscillates without settling. $\sin\left(\frac{1}{x}\right)$ as $x \to 0$ swings between $-1$ and $1$ infinitely often however close $x$ gets to $0$, so no single value is approached.

Estimating limits from graphs and tables

On a graph, follow the curve towards $x = c$ from each side and read the height it is heading for. An open circle at $x = c$ means the function is not defined or takes a different value there, and the limit is the height of the hole. A graph drawn at a fixed scale can hide behaviour, such as a very fast oscillation or a narrow spike, so a graph supports an estimate and does not prove one.

From a table, choose values of $x$ closer and closer to $c$ on both sides and watch the outputs. For $\frac{e^x - 1}{x}$ near $0$:

The outputs close in on $1$ from both sides, so the limit is estimated as $1$. A table with values from only one side can only estimate the one-sided limit. A trend across a few values is evidence and not a guarantee.

Limit laws

If $\lim_{x \to c} f(x) = L$ and $\lim_{x \to c} g(x) = M$ both exist as real numbers, then

For a composite function, $\lim_{x \to c} f(g(x)) = f\left(\lim_{x \to c} g(x)\right)$ when $f$ is continuous at that limit of $g$. Polynomials, rational functions (where the denominator is not zero), $\sqrt[n]{x}$, $\sin x$, $\cos x$, $e^x$ and $\ln x$ are continuous on their domains, so their limits at a point of the domain are found by substituting. For instance $\lim_{x \to 2} \frac{x^2 + 1}{x + 3} = \frac{5}{5} = 1$.

When substitution gives $\frac{0}{0}$

Substituting $x = c$ can give $\frac{0}{0}$. This is an indeterminate form and says nothing about the limit yet: the expression has to be rewritten first. The rewriting works because limits ignore the value at $c$ itself, so a common factor that vanishes at $c$ can be cancelled.

Factoring

Multiplying by a conjugate

When a square root sits in the numerator or denominator, multiplying the top and bottom by the conjugate removes it:

Combining fractions

Trigonometric identities

Two limits are used repeatedly:

They extend to other multiples of the variable by rescaling. For $\lim_{x \to 0} \frac{\sin 3x}{x}$, write $\frac{\sin 3x}{x} = 3 \cdot \frac{\sin 3x}{3x}$, and since $3x \to 0$ as $x \to 0$, the limit is $3 \cdot 1 = 3$.

If substitution gives a nonzero number over zero, such as $\frac{5}{0}$, the expression is not indeterminate. The function is unbounded near $c$ and the limit is either infinite or does not exist.

The squeeze theorem

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

It is used when the function is trapped between two simpler ones. To find $\lim_{x \to 0} x^2 \sin\left(\frac{1}{x}\right)$, note that $-1 \le \sin\left(\frac{1}{x}\right) \le 1$ for $x \ne 0$, and multiplying by $x^2 \ge 0$ gives

Both $-x^2$ and $x^2$ tend to $0$ as $x \to 0$, so the limit is $0$. Without the factor $x^2$, $\sin\left(\frac{1}{x}\right)$ has no limit at $0$, so the squeeze is what tames the oscillation.

The two trigonometric limits above are themselves proved by squeezing, using a comparison of areas in a unit circle.

Choosing a method

Try substitution first. If the result is a number, it is the limit, provided the function is continuous there. If it is $\frac{0}{0}$, look for a factor to cancel, a conjugate to multiply by, a fraction to combine or a trigonometric identity to apply. If the expression involves a bounded oscillating factor multiplied by something that tends to zero, consider the squeeze theorem. For piecewise functions, or at a point where the formula changes, work out each one-sided limit separately. Where the form is $\frac{0}{0}$ or $\frac{\infty}{\infty}$ and the expression is differentiable, L'Hospital's rule gives another route once derivatives are available.