Limits and Continuity: The Foundation of Calculus

The limit is the conceptual move that allows calculus to talk about instantaneous rates of change and continuous accumulations. Before the limit, there was no way to make precise the idea of the slope at a single point — you could only talk about average slopes over intervals. The limit closes the gap, and calculus follows.

This article covers what limits intuitively mean, the formal epsilon-delta definition, how to compute limits in practice, L'Hôpital's rule for indeterminate forms, continuity and its types, infinite limits and limits at infinity, why this material matters for the rest of calculus, and common pitfalls.

Why limits exist as a concept

Consider the question: what is the slope of the curve $f(x)=x^2$ at the point $x=1$?

Earlier mathematics could ask: what is the slope between $x=1$ and $x=2$? That is computable: $\frac{f(2)-f(1)}{2-1}=\frac{4-1}{1}=3$. This is the slope of the secant line between the two points — an average slope over an interval.

But what is the slope at the single point $x=1$? You cannot compute $\frac{f(1)-f(1)}{1-1}=\frac{0}{0}$, which is undefined. The problem is that at a single point, there is no interval over which to compute a slope.

The limit resolves the difficulty. Consider the slope between $x=1$ and $x=1+h$ for small $h$:

$\frac{f(1+h)-f(1)}{h}=\frac{(1+h{)}^2-1}{h}=\frac{2h+h^2}{h}=2+h$

As $h$ shrinks to zero, this expression approaches 2. The limit is 2 — and we say the slope of $f(x)=x^2$ at $x=1$ is 2.

This is the central move. The limit allows us to talk about what an expression approaches as a variable approaches a value, even when the expression is undefined at that value. The slope at a point is the limit of the average slope.

The intuitive definition

We write ${\lim }_{x\to a}f(x)=L$ to mean: as $x$ gets arbitrarily close to $a$ (but not equal to $a$), $f(x)$ gets arbitrarily close to $L$.

This intuitive definition is enough for most purposes. The technical move is to make "arbitrarily close" precise.

Example 1: ${\lim }_{x\to 2}(x^2+3)=?$

As $x$ gets close to 2, $x^2$ gets close to 4, and $x^2+3$ gets close to 7. So the limit is 7.

This works because the function $x^2+3$ is continuous at $x=2$ — its value at $x=2$ equals the limit as $x\to 2$. For continuous functions, computing limits is just substituting.

Example 2: ${\lim }_{x\to 1}\frac{x^2-1}{x-1}=?$

Direct substitution gives $\frac{0}{0}$ — undefined. But you can factor: $\frac{x^2-1}{x-1}=\frac{(x-1)(x+1)}{x-1}=x+1$ for all $x\ne 1$. The limit as $x\to 1$ is $1+1=2$.

Even though the original expression is not defined at $x=1$, the limit exists and equals 2. The function "wants to be" 2 at $x=1$.

The formal epsilon-delta definition

The intuitive definition uses informal words ("arbitrarily close"). The 19th-century mathematicians Cauchy and Weierstrass made it precise:

${\lim }_{x\to a}f(x)=L$ means: for every $\epsilon >0$, there exists $\delta >0$ such that whenever $0<|x-a|<\delta$, we have $|f(x)-L|<\epsilon$.

In plain English: no matter how close (how small an $\epsilon$) you want $f(x)$ to be to $L$, you can find a neighborhood around $a$ (a $\delta$) such that all $x$ in that neighborhood (except $a$ itself) produce $f(x)$ within $\epsilon$ of $L$.

This is the rigorous definition. Almost all of analysis (the rigorous foundation of calculus) is built on it.

For practical computation, you almost never use the epsilon-delta definition directly. You use the limit laws derived from it.

Limit laws

If ${\lim }_{x\to a}f(x)=L$ and ${\lim }_{x\to a}g(x)=M$, then:

• Sum: ${\lim }_{x\to a}(f(x)+g(x))=L+M$
• Product: ${\lim }_{x\to a}(f(x)\cdot g(x))=L\cdot M$
• Quotient: ${\lim }_{x\to a}\frac{f(x)}{g(x)}=\frac{L}{M}$, provided $M\ne 0$
• Constant multiple: ${\lim }_{x\to a}(c\cdot f(x))=c\cdot L$
• Power: ${\lim }_{x\to a}f(x{)}^n=L^n$ for positive integers $n$

For polynomial functions, you can always compute the limit by substitution: ${\lim }_{x\to a}p(x)=p(a)$. The limit laws guarantee it.

For rational functions $\frac{p(x)}{q(x)}$, you can substitute if $q(a)\ne 0$. If $q(a)=0$, you need other techniques.

Computing limits: techniques

When direct substitution fails (typically because of $\frac{0}{0}$ or $\frac{\infty }{\infty }$), use one of these.

Technique 1 — Algebraic simplification.

Factor and cancel, rationalize, or otherwise rewrite the expression to remove the problematic point.

Example 3: ${\lim }_{x\to 0}\frac{\sqrt{x+4}-2}{x}$

Multiply numerator and denominator by the conjugate $\sqrt{x+4}+2$:

$\frac{(\sqrt{x+4}-2)(\sqrt{x+4}+2)}{x(\sqrt{x+4}+2)}=\frac{(x+4)-4}{x(\sqrt{x+4}+2)}=\frac{x}{x(\sqrt{x+4}+2)}=\frac{1}{\sqrt{x+4}+2}$

Now substitute $x=0$: $\frac{1}{\sqrt{4}+2}=\frac{1}{4}$.

Technique 2 — L'Hôpital's rule.

If the limit produces $\frac{0}{0}$ or $\frac{\infty }{\infty }$, you may differentiate both numerator and denominator and try the limit again:

${\lim }_{x\to a}\frac{f(x)}{g(x)}={\lim }_{x\to a}\frac{f^{\prime }(x)}{g^{\prime }(x)}$

provided the right-hand side exists.

Example 4: ${\lim }_{x\to 0}\frac{\sin x}{x}$.

This is the classic. Direct substitution gives $\frac{0}{0}$. Apply L'Hôpital:

${\lim }_{x\to 0}\frac{\cos x}{1}=\cos (0)=1$

So ${\lim }_{x\to 0}\frac{\sin x}{x}=1$.

(Note: a careful proof of this limit actually uses geometric arguments, not L'Hôpital, because L'Hôpital's rule depends on knowing $\frac{d}{dx}(\sin x)=\cos x$, which is itself derived from this limit. The circular reasoning is broken in proper treatments. But for computation, L'Hôpital works.)

Example 5: ${\lim }_{x\to \infty }\frac{x^2}{e^x}$.

Direct gives $\frac{\infty }{\infty }$. Apply L'Hôpital: ${\lim }_{x\to \infty }\frac{2x}{e^x}$. Still $\frac{\infty }{\infty }$. Apply again: ${\lim }_{x\to \infty }\frac{2}{e^x}=0$.

The limit is 0. Exponentials grow faster than polynomials.

Technique 3 — Special limits memorized.

A handful of limits come up repeatedly and should be memorized:

${\lim }_{x\to 0}\frac{\sin x}{x}=1$ ${\lim }_{x\to 0}\frac{1-\cos x}{x}=0$ ${\lim }_{x\to 0}\frac{e^x-1}{x}=1$ ${\lim }_{x\to 0}\frac{\ln (1+x)}{x}=1$ ${\lim }_{x\to \infty }{\left(1+\frac{1}{x}\right)}^x=e$

Continuity

A function $f$ is continuous at $a$ if:

1. $f(a)$ is defined.
2. ${\lim }_{x\to a}f(x)$ exists.
3. ${\lim }_{x\to a}f(x)=f(a)$.

All three conditions matter. If any fails, $f$ is not continuous at $a$.

Types of discontinuity:

• Removable discontinuity: ${\lim }_{x\to a}f(x)$ exists but does not equal $f(a)$ (or $f(a)$ is undefined). Example: $f(x)=\frac{x^2-1}{x-1}$ at $x=1$. The limit is 2, but the function is undefined there. Can be "fixed" by redefining $f(1)=2$.

• Jump discontinuity: the function "jumps" — left limit and right limit exist but differ. Example: $f(x)=\frac{|x|}{x}$ at $x=0$. Left limit is $-1$, right limit is $+1$.

• Infinite discontinuity: the function goes to $\pm \infty$ at the point. Example: $f(x)=\frac{1}{x}$ at $x=0$.

Continuous functions are well-behaved. The standard theorems of calculus — the Intermediate Value Theorem, the Extreme Value Theorem, the Mean Value Theorem — all require continuity. Continuity guarantees that small changes in input produce small changes in output.

Most functions you encounter — polynomials, exponentials, sines, cosines, $\sqrt{x}$ (where defined), logarithms (where defined) — are continuous on their domains.

One-sided limits

Sometimes a function behaves differently on the two sides of a point. The one-sided limits capture this.

${\lim }_{x\to a^{-}}f(x)=L$ means $f(x)\to L$ as $x$ approaches $a$ from the left (from values less than $a$).

${\lim }_{x\to a^{+}}f(x)=L$ means $f(x)\to L$ as $x$ approaches $a$ from the right (from 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 6: $f(x)=|x|/x$ at $x=0$.

${\lim }_{x\to 0^{-}}\frac{|x|}{x}=\frac{-x}{x}=-1$ (since $|x|=-x$ for $x<0$). ${\lim }_{x\to 0^{+}}\frac{|x|}{x}=\frac{x}{x}=1$.

The two-sided limit does not exist.

Infinite limits and limits at infinity

Infinite limits: ${\lim }_{x\to a}f(x)=\infty$ means $f(x)$ grows without bound as $x\to a$. Example: ${\lim }_{x\to 0^{+}}\frac{1}{x}=\infty$.

Limits at infinity: ${\lim }_{x\to \infty }f(x)=L$ means $f(x)\to L$ as $x$ grows without bound. Example: ${\lim }_{x\to \infty }\frac{1}{x}=0$.

Combine: ${\lim }_{x\to \infty }f(x)=\infty$ means $f$ grows without bound as $x$ grows without bound. Example: ${\lim }_{x\to \infty }{x}^2=\infty$.

For rational functions ${\lim }_{x\to \infty }\frac{p(x)}{q(x)}$:

• If degree of $p$ < degree of $q$: limit is 0.
• If degree of $p$ = degree of $q$: limit is the ratio of leading coefficients.
• If degree of $p$ > degree of $q$: limit is $\pm \infty$ (sign depends on leading coefficients).

Why this matters

Three reasons.

Reason 1 — The derivative is defined as a limit.

The derivative of $f$ at $x$ is

$f^{\prime }(x)={\lim }_{h\to 0}\frac{f(x+h)-f(x)}{h}$

Without limits, this is not even a well-defined expression. Every derivative computation, every differentiation rule, descends from limits.

Reason 2 — The integral is defined as a limit.

The definite integral ${\int }_a^{b}f(x)dx$ is the limit of Riemann sums as the partition becomes infinitely fine. Without limits, no integration.

Reason 3 — Continuity is the basis of analysis.

The Intermediate Value Theorem, the Extreme Value Theorem, uniform convergence, differential equations — all assume continuity. Continuity is the precondition for most of the powerful theorems in mathematics.

Common pitfalls

Pitfall 1 — Confusing limit and value.

${\lim }_{x\to a}f(x)$ is about what $f(x)$ approaches as $x$ approaches $a$, not about $f(a)$. The two can differ — and when they differ, the function is not continuous.

Pitfall 2 — Plugging in numerically and reading off.

Trying ${\lim }_{x\to 0}\frac{\sin x}{x}$ by computing $\sin (0.0001)/0.0001$ on a calculator gives an approximation, not the answer. The exact answer is 1, but you cannot conclude this from one numerical value. The argument requires algebra or L'Hôpital.

Pitfall 3 — Applying L'Hôpital where it does not apply.

L'Hôpital only applies to $\frac{0}{0}$ or $\frac{\infty }{\infty }$ forms. If the limit is $\frac{1}{0}$ or $\frac{0}{1}$, L'Hôpital gives wrong answers. Check the form before applying.

Pitfall 4 — Assuming the limit always exists.

Some limits do not exist. ${\lim }_{x\to 0}\sin (1/x)$ does not exist — the function oscillates infinitely fast near zero. There is no value $L$ that $\sin (1/x)$ approaches.