Honors Pre-Calculus Unit 12 Review: Introduction to Calculus

A limit tells you the value $f(x)$ approaches as $x$ gets close to $a$. You write it as $\lim_{x \to a} f(x) = L$. What happens exactly at $x = a$ doesn't matter, and the function doesn't even need to be defined there.

The numerical approach uses a table. Pick inputs closer and closer to $a$ from both sides and watch where the outputs settle.

Worked example. Estimate $\lim_{x \to 3} \frac{x^2 - 9}{x - 3}$. From the left, $x = 2.9$ gives $5.9$ and $x = 2.99$ gives $5.99$. From the right, $x = 3.1$ gives $6.1$ and $x = 3.01$ gives $6.01$. Both sides head toward 6, so the limit is 6, even though plugging in 3 gives $\frac{0}{0}$.

The graphical approach works the same way. Trace the curve toward $x = a$ from the left, then from the right, and read the $y$-value each side approaches. An open circle at that spot doesn't change the limit.

If the two sides approach different values, the two-sided limit does not exist (DNE). If the outputs grow without bound, as with $\frac{1}{x^2}$ near 0, you can write $\lim_{x \to 0} \frac{1}{x^2} = \infty$ to describe the behavior, but there is no finite limit.

Limit laws let you calculate limits exactly instead of estimating them. If $\lim_{x \to a} f(x)$ and $\lim_{x \to a} g(x)$ both exist as finite numbers, you can split a limit across sums, differences, products, constant multiples, and positive integer powers. Root laws apply where the root expression is defined. You can also split a limit across quotients when the denominator's limit isn't 0.

For polynomials, direct substitution always works. For example, $\lim_{x \to 2}(3x^2 - 2x + 1) = 12 - 4 + 1 = 9$. Rational functions work the same way as long as the denominator isn't 0 at $a$.

If substitution gives $\frac{0}{0}$, it doesn't determine the limit. Factoring and canceling or multiplying by a conjugate may help, but the limit can also fail to exist.

Worked example. Find $\lim_{x \to 4} \frac{\sqrt{x} - 2}{x - 4}$. Substitution gives $\frac{0}{0}$. Multiply the top and bottom by $\sqrt{x} + 2$. The numerator becomes $x - 4$, so the expression simplifies to $\frac{1}{\sqrt{x} + 2}$. Now substitute to get $\frac{1}{2 + 2} = \frac{1}{4}$.

For a rational function, if the numerator is nonzero and the denominator is 0 at $x = a$, there is a vertical asymptote at $x = a$. Check each side to see whether the function heads to $\infty$ or $-\infty$.

For limits as $x \to \infty$ of rational functions, divide every term by the highest power of $x$ in the denominator. For example, $\lim_{x \to \infty} \frac{3x^2 + 2x}{5x^2 - 1} = \frac{3}{5}$.

At an interior point $x = a$, a function is continuous when three conditions all hold. First, $f(a)$ is defined. Second, $\lim_{x \to a} f(x)$ exists. Third, $\lim_{x \to a} f(x) = f(a)$. At an endpoint of an interval in the domain, use the one-sided limit from within the domain.

Polynomials, $e^x$, $\sin x$, and $\cos x$ are continuous everywhere. Rational functions are continuous everywhere in their domain. The places to check are zeros of a denominator, even roots of negative numbers, and the break points of piecewise functions.

Worked example. Find $k$ so that $f(x) = x + 1$ for $x < 2$ and $f(x) = kx$ for $x \ge 2$ is continuous. The left-hand limit at 2 is $2 + 1 = 3$. The right-hand limit and $f(2)$ both equal $2k$. Setting $2k = 3$ gives $k = 1.5$.

Continuity also lets you use the Intermediate Value Theorem (IVT). If $f$ is continuous on $[a, b]$, it hits every value between $f(a)$ and $f(b)$. For example, $f(x) = x^3 - x - 1$ has $f(1) = -1$ and $f(2) = 5$, so it has a zero somewhere between 1 and 2.

The derivative $f'(a)$ is the slope of the curve at the single point $x = a$. It comes from the slope of a secant line through $(a, f(a))$ and a nearby point $(a + h, f(a + h))$, with $h$ shrinking to 0:

$f'(a) = \lim_{h \to 0} \frac{f(a+h) - f(a)}{h}$

Worked example. Find the tangent line to $f(x) = x^2$ at $x = 3$. The difference quotient is $\frac{(3+h)^2 - 9}{h} = \frac{6h + h^2}{h} = 6 + h$. As $h \to 0$, this approaches 6, so $f'(3) = 6$. The point is $(3, 9)$. In point-slope form, $y - 9 = 6(x - 3)$, which simplifies to $y = 6x - 9$.

The power rule is a shortcut for the same work. The derivative of $x^n$ is $nx^{n-1}$. Constants multiply through, and you differentiate sums term by term. So $f(x) = 2x^3 - 3x + 1$ has $f'(x) = 6x^2 - 3$.

As a rate, the derivative measures instantaneous change. If $s(t)$ is position, then $v(t) = s'(t)$ is velocity and $a(t) = v'(t)$ is acceleration.

A function can be continuous without being differentiable. The function $|x|$ has a sharp corner at 0, where the slope is $-1$ from the left and $1$ from the right, so $f'(0)$ doesn't exist.