Limits & Continuity

Before studying limits, you need fluency with the functions that calculus operates on. Every model in science begins with choosing the right function family.

Linear functions $f(x) = mx + b$ model constant rates of change. Their graphs are straight lines, and their simplicity is what derivatives approximate locally — the tangent line is a linear approximation.

Power functions $f(x) = x^n$ produce polynomial behavior. Quadratics model projectile motion, cubics model inflection-style curves. Polynomials are the most 'well-behaved' functions: continuous, differentiable, no surprises.

Exponential functions $f(x) = a^x$ (especially $e^x$) model growth and decay — population, radioactivity, compound interest. Their defining property: the rate of growth is proportional to the current value. This is why $e^x$ is its own derivative.

Logarithmic functions $f(x) = \log_a(x)$ are the inverses of exponentials. $\ln x$ grows without bound, but incredibly slowly. It takes center stage in integration ($\int 1/x\,dx = \ln|x| + C$).

Trigonometric functions $\sin x$, $\cos x$, $\tan x$ model periodic phenomena — waves, oscillations, circular motion. Their derivatives cycle through each other, and their limits (especially $\lim_{x\to 0} \sin x / x = 1$) are foundational.

Inverse functions reverse input and output. If $f(x) = e^x$, then $f^{-1}(x) = \ln x$. For inverses to exist, $f$ must be one-to-one. The derivative of an inverse function satisfies $(f^{-1})'(x) = \frac{1}{f'(f^{-1}(x))}$.

Piecewise functions are defined by different rules on different intervals. They are the natural setting for studying one-sided limits, continuity, and differentiability — the exact concepts coming next.

Imagine plugging in values of $x$ that get closer and closer to $a$. For $f(x) = (x^2-1)/(x-1)$, try $x=0.9, 0.99, 0.999, 1.001, 1.01, 1.1$. You'll see the outputs cluster around $2$, even though $f(1)$ is undefined.

That clustering is the limit. Formally, $\lim_{x\to 1} \frac{x^2-1}{x-1} = 2$ because the outputs can be made as close to $2$ as we like by choosing $x$ close enough to $1$.

The limit cares about the neighborhood around a point, not the point itself. This is why limits are useful: they let us analyze behavior at places where the function might break.

The formal $\varepsilon$-$\delta$ definition makes this precise: $\lim_{x\to a} f(x) = L$ means for every $\varepsilon > 0$, there exists a $\delta > 0$ such that if $0 < |x - a| < \delta$ then $|f(x) - L| < \varepsilon$. You pick a tolerance $\varepsilon$ for the output, and the definition guarantees a tolerance $\delta$ for the input that works.

The left-hand limit $\lim_{x\to a^-} f(x)$ only uses $x$-values less than $a$ (approaching from the left). The right-hand limit $\lim_{x\to a^+} f(x)$ only uses values greater than $a$.

The two-sided limit $\lim_{x\to a} f(x)$ exists if and only if both one-sided limits exist and are equal. This is the bridge between one-sided behavior and a 'full' limit.

Example: for $f(x) = |x|/x$, we have $\lim_{x\to 0^-} f(x) = -1$ and $\lim_{x\to 0^+} f(x) = 1$. Since they disagree, the two-sided limit at $0$ does not exist.

One-sided limits are essential for piecewise functions, absolute value expressions, and functions with jumps. They also arise naturally at domain boundaries — for instance, $\lim_{x\to 0^+} \ln x = -\infty$, and asking about the left-hand limit doesn't make sense since $\ln x$ is undefined for $x < 0$.

Always try direct substitution first. If $f$ is continuous at $a$, then $\lim_{x\to a} f(x) = f(a)$. You're done. This is the simplest and fastest method.

Polynomials, rational functions (away from zeros of the denominator), $e^x$, $\ln x$ (for $x>0$), $\sin x$, $\cos x$, and $\tan x$ (away from asymptotes) are all continuous on their domains. Compositions, sums, and products of continuous functions are also continuous.

Direct substitution fails when you get an undefined expression like $0/0$ or $\infty/\infty$. These are called indeterminate forms, and they signal that the function's behavior at that point is more complex — the limit might exist, but you need algebraic techniques or L'Hôpital's Rule to find it.

If substitution gives something like $5/0$ (nonzero over zero), the limit doesn't exist as a finite number — the function blows up. This typically signals a vertical asymptote.

The expression $0/0$ is called indeterminate because the limit could be anything: $0$, $5$, $\infty$, or it might not exist. You must simplify.

Factor and cancel: if the numerator and denominator share a common factor like $(x-a)$, cancel it and try substitution again.

Rationalize: when you see a square root, multiply by the conjugate. For example, $\frac{\sqrt{x+1}-1}{x}$ times $\frac{\sqrt{x+1}+1}{\sqrt{x+1}+1}$ simplifies the radical.

Combine fractions: if the expression has two fractions being subtracted, find a common denominator to combine them into a single fraction.

Use trig identities: rewrite expressions using $\sin^2 x + \cos^2 x = 1$, double-angle formulas, or the identity $1-\cos x = 2\sin^2(x/2)$ to simplify.

These special limits appear constantly and should be memorized:

$\lim_{x\to 0} \frac{\sin x}{x} = 1$. This is the most important trig limit in calculus. It's proven using the Squeeze Theorem.

$\lim_{x\to 0} \frac{1-\cos x}{x} = 0$. You can derive this by multiplying by $\frac{1+\cos x}{1+\cos x}$ and using the first limit.

$\lim_{x\to 0} \frac{\tan x}{x} = 1$. This follows from $\frac{\tan x}{x} = \frac{\sin x}{x} \cdot \frac{1}{\cos x}$.

Generalization: $\lim_{x\to 0} \frac{\sin(kx)}{x} = k$ for any constant $k$. Factor out the $k$ or substitute $u=kx$.

Another useful limit: $\lim_{x\to 0} \frac{e^x - 1}{x} = 1$. This shows up in many exponential limit problems.

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

The function $f$ is 'squeezed' between two functions that both approach the same value, so $f$ has no choice but to approach that value too.

Classic example: $\lim_{x\to 0} x^2 \sin(1/x)$. We know $-x^2 \leq x^2\sin(1/x) \leq x^2$, and both $-x^2$ and $x^2$ approach $0$, so the limit is $0$.

The Squeeze Theorem is how we rigorously prove $\lim_{x\to 0} \frac{\sin x}{x} = 1$ using geometry of the unit circle.

A function is continuous at $x=a$ when three conditions all hold: (1) $f(a)$ is defined, (2) $\lim_{x\to a} f(x)$ exists, and (3) $\lim_{x\to a} f(x) = f(a)$.

If any condition fails, we have a discontinuity. Types of discontinuity:

Removable discontinuity (hole): the limit exists but $f(a)$ is either undefined or disagrees with the limit. Example: $f(x) = (x^2-1)/(x-1)$ at $x=1$.

Jump discontinuity: the left and right limits exist but are different. Example: $f(x) = \lfloor x \rfloor$ (floor function) at every integer.

Infinite discontinuity: the function blows up to $\pm\infty$. Example: $f(x) = 1/x$ at $x=0$.

A function continuous on a closed interval $[a,b]$ is guaranteed to hit every $y$-value between $f(a)$ and $f(b)$ (Intermediate Value Theorem). This is used to prove equations have solutions.

For a piecewise function, you must check the left-hand and right-hand limits separately at each boundary point. Each side of the boundary uses a different formula.

Use the piece that applies for $x < a$ to compute $\lim_{x\to a^-}$, and the piece that applies for $x > a$ to compute $\lim_{x\to a^+}$.

If $\lim_{x\to a^-} f(x) = \lim_{x\to a^+} f(x) = L$, then $\lim_{x\to a} f(x) = L$. For continuity, you additionally need $f(a) = L$ — the actual function value at the boundary must match.

Three outcomes at a boundary: (1) both sides match and equal $f(a)$ — continuous; (2) both sides match but $f(a)$ differs — removable discontinuity; (3) the sides disagree — jump discontinuity.

An infinite limit like $\lim_{x\to a} f(x) = \infty$ means the outputs grow without bound as $x$ approaches $a$. The limit technically does not exist as a real number, but we write $\infty$ to describe the behavior.

Vertical asymptotes occur where the denominator approaches $0$ but the numerator does not. The graph shoots upward or downward near these points.

Check signs carefully: $\lim_{x\to 0^+} 1/x = +\infty$ but $\lim_{x\to 0^-} 1/x = -\infty$. The one-sided behavior can differ, so always examine both sides. If the two sides have opposite signs, the graph has an asymptote going in both directions.

For rational functions, factor the denominator to find all vertical asymptotes. Cancel common factors first — those give removable discontinuities (holes), not asymptotes.

For functions like $\ln x$, $\tan x$, and $\csc x$, the asymptotes come from the nature of the function rather than a denominator. For example, $\tan x$ has vertical asymptotes at $x = \pm \pi/2, \pm 3\pi/2, \ldots$

Limits as $x\to \infty$ describe long-run behavior. A horizontal asymptote $y=L$ means $\lim_{x\to \infty} f(x) = L$ or $\lim_{x\to -\infty} f(x) = L$.

For rational functions $\frac{p(x)}{q(x)}$, compare the degrees: if $\deg(p) < \deg(q)$, the limit is $0$. If $\deg(p) = \deg(q)$, the limit is the ratio of leading coefficients. If $\deg(p) > \deg(q)$, the limit is $\pm\infty$ (no horizontal asymptote).

Technique: divide every term by the highest power of $x$ in the denominator. As $x\to \infty$, terms like $1/x$, $1/x^2$, etc. all go to $0$.

For expressions with radicals, factor out the dominant term. Example: $\lim_{x\to \infty} \frac{\sqrt{4x^2+1}}{x} = \lim_{x\to \infty} \frac{x\sqrt{4+1/x^2}}{x} = 2$.

When direct substitution gives $0/0$ or $\infty/\infty$, L'Hôpital's Rule says: $\lim_{x\to a} \frac{f(x)}{g(x)} = \lim_{x\to a} \frac{f'(x)}{g'(x)}$, provided the right-hand limit exists.

Important: differentiate the numerator and denominator separately — not as a single fraction. This is not the quotient rule.

After applying the rule once, check whether the new limit is still indeterminate. If it is, apply L'Hôpital's Rule again. Repeat until you reach a form where substitution gives a definite value. For example, $\lim_{x\to 0} \frac{e^x - 1 - x}{x^2}$ requires two applications: the first gives $\frac{e^x - 1}{2x}$ (still $0/0$), and the second gives $\frac{e^x}{2} \to \frac{1}{2}$.

Before each application, always verify the form is still $0/0$ or $\infty/\infty$. If it isn't — for instance, if the denominator approaches a nonzero constant — stop and evaluate directly. Applying L'Hôpital to a non-indeterminate form produces wrong answers.

For other indeterminate forms like $0 \cdot \infty$, $\infty - \infty$, $1^\infty$, $0^0$, or $\infty^0$, rewrite the expression into a $0/0$ or $\infty/\infty$ form first, then apply L'Hôpital.

Example: for $\lim_{x\to 0^+} x\ln x$ ($0 \cdot (-\infty)$ form), rewrite as $\lim_{x\to 0^+} \frac{\ln x}{1/x}$ ($-\infty / \infty$ form) and apply L'Hôpital.