Limits & Continuity

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 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.

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.

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

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.

Direct substitution fails when you get an undefined expression like $0/0$. That's when you need the techniques below.

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.

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 also need $f(a) = L$.

Example: if $f(x) = x^2$ for $x < 2$ and $f(x) = 3x-2$ for $x \geq 2$, then $\lim_{x\to 2^-} f(x) = 4$ and $\lim_{x\to 2^+} f(x) = 4$, so the limit exists and equals $4$. Since $f(2)=4$, the function is continuous at $x=2$.

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.

Check signs carefully: $\lim_{x\to 0^+} 1/x = +\infty$ but $\lim_{x\to 0^-} 1/x = -\infty$. The one-sided behavior can differ.

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

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. Do not use the quotient rule.

You may need to apply the rule multiple times if the result is still indeterminate.

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.

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.