Limits and Continuity

Use limits to describe nearby behavior without confusing approaching a value with reaching it. Connect one-sided limits and continuity to the limiting slope that defines a derivative.

A limit describes the value that a function approaches as its input approaches a chosen point. The function need not equal that value at the point, or even be defined there. Limits therefore separate nearby behavior from direct evaluation and provide the precise foundation for continuity and derivatives.

A Limit Describes Nearby Behavior

The notation:

\lim_{x\to a}f(x)=L

Read this as: “as $x$ approaches $a$ without needing to equal it, the values of $f(x)$ approach $L$ .” The notation describes nearby behavior, not necessarily the value assigned at $a$ .

is read as “the limit of $f(x)$ as $x$ approaches $a$ is $L$ .” It means that $f(x)$ can be made as close to $L$ as desired by taking $x$ sufficiently close to $a$ , without requiring $x=a$ .

Consider:

f(x)=\frac{x^2-1}{x-1}.

Direct substitution at $x=1$ gives $0/0$ , so the fraction is undefined there. For every $x\ne1$ , however:

\frac{x^2-1}{x-1} =\frac{(x-1)(x+1)}{x-1} =x+1.

Nearby values therefore approach $2$ as $x$ approaches $1$ :

\lim_{x\to1}\frac{x^2-1}{x-1}=2.

The cancellation does not define the original fraction at $x=1$ . It reveals that the fraction and $x+1$ have the same values everywhere near that point.

Exercise: Evaluate a removable limit

What is:

\lim_{x\to2}\frac{x^2-4}{x-2}?

Check the limit

Compute it first, then check your number.

HintFactor the numerator

Use $x^2-4=(x-2)(x+2)$ , then simplify for $x\ne2$ .

SolutionSimplify the nearby expression

For $x\ne2$ :

\frac{x^2-4}{x-2} =\frac{(x-2)(x+2)}{x-2} =x+2.

As $x$ approaches $2$ , this nearby expression approaches $2+2=4$ . The limit is $4$ , even though the original fraction is undefined at exactly $x=2$ .

Approaching Is Not the Same as Evaluating

Three quantities must be kept separate:

the nearby limit $\lim_{x\to a}f(x)$ ;
the point value $f(a)$ ;
whether the point value is defined at all.

They often agree, but they need not. Define:

g(x)= \begin{cases} x+1,&x\ne1,\\ 10,&x=1. \end{cases}

The nearby outputs still approach $2$ , so:

\lim_{x\to1}g(x)=2,

while $g(1)=10$ . Changing one isolated point does not change the limit because the limit concerns values arbitrarily close to the point rather than the value at the point itself.

Approach a missing point from both sides

The expression is not evaluated at x = 1. Instead, the table samples inputs that move progressively closer to 1 from below and above.

Ready to run.

Numerical tables can suggest a limit, but they do not prove it. Samples inspect only finitely many inputs, and rounding can hide behavior. Algebraic simplification, known limit laws, inequalities, or other arguments establish the mathematical result.

Exercise: Separate a limit from a point value

Suppose:

h(x)= \begin{cases} 2x,&x\ne3,\\ -1,&x=3. \end{cases}

What is $\lim_{x\to3}h(x)$ ?

Check the limit

Compute it first, then check your number.

HintInspect nearby inputs

When $x$ is close to but unequal to $3$ , the active rule is $h(x)=2x$ .

SolutionUse the nearby rule

Every nearby input other than $3$ uses $h(x)=2x$ . Therefore:

\lim_{x\to3}h(x) =\lim_{x\to3}2x =6.

The point value is $h(3)=-1$ , so the limit exists but does not equal the function value.

One-Sided Limits Must Agree

A two-sided limit approaches the point from both directions. The left-hand limit uses smaller inputs:

\lim_{x\to a^-}f(x),

and the right-hand limit uses larger inputs:

\lim_{x\to a^+}f(x).

The two-sided limit exists exactly when both one-sided limits exist and are equal.

For the step function:

s(x)= \begin{cases} 0,&x<0,\\ 1,&x\ge0, \end{cases}

we have:

\lim_{x\to0^-}s(x)=0, \qquad \lim_{x\to0^+}s(x)=1.

Since the approach values disagree, $\lim_{x\to0}s(x)$ does not exist. The assigned value $s(0)=1$ cannot repair that disagreement.

Exercise: Diagnose a jump

A function approaches $3$ from the left of $x=5$ and $7$ from the right. Does its two-sided limit at $5$ exist?

Choose the supported conclusion

No. The one-sided limits disagree.

Yes. The limit is the midpoint, 5.

Yes. The right-hand limit determines it.

Select one choice, then check.

HintRequire one approach value

A two-sided limit exists only when the left- and right-hand limits agree.

SolutionCompare both directions

The left-hand limit is $3$ and the right-hand limit is $7$ . Since they are unequal, there is no single number approached from both sides. The two-sided limit therefore does not exist, regardless of the function's value at $5$ .

Limit Laws and Direct Substitution

When the relevant limits exist, limits respect ordinary algebra. For example:

\lim_{x\to a}(f(x)+g(x)) =\lim_{x\to a}f(x)+\lim_{x\to a}g(x),

and similarly for differences, products, constant multiples, and quotients whose limiting denominator is nonzero.

Polynomials are continuous everywhere, so their limits can be found by direct substitution. Rational functions are continuous wherever their denominator is nonzero. Thus:

\lim_{x\to2}(x^2+3x-1) =2^2+3(2)-1 =9.

An indeterminate expression such as $0/0$ is not the value of a limit. It is a signal that direct substitution has not decided the question. Factoring may reveal a removable hole, but other $0/0$ expressions can have a different finite limit, no finite limit, or unbounded behavior.

Continuity Connects the Limit and the Value

A function is continuous at $x=a$ when all three conditions hold:

$f(a)$ is defined;
$\lim_{x\to a}f(x)$ exists;
$\lim_{x\to a}f(x)=f(a)$ .

A removable discontinuity fails the first or third condition even though the nearby limit exists. It can be repaired by assigning the missing point the limiting value. A jump discontinuity fails the second condition because the one-sided limits disagree. An unbounded discontinuity occurs when nearby outputs grow without a finite bound, as $1/x$ does near $0$ .

Continuity does not mean flat, straight, or slowly changing. The exponential function is continuous but can grow rapidly. Informally, continuity says that nearby inputs produce nearby outputs. The three conditions above state exactly how that idea applies at one point.

Exercise: Check continuity at a point

A function satisfies:

f(2)=5, \qquad \lim_{x\to2^-}f(x)=5, \qquad \lim_{x\to2^+}f(x)=5.

Is $f$ continuous at $x=2$ ?

Choose the continuity conclusion

Yes. Both one-sided limits and the point value agree.

No. A limit can never equal a function value.

It cannot be decided without a formula for every input.

Select one choice, then check.

HintBuild the two-sided limit

Equal one-sided limits give $\lim_{x\to2}f(x)=5$ ; compare that result with $f(2)$ .

SolutionVerify all three conditions

The function value $f(2)=5$ is defined. Both one-sided limits equal $5$ , so the two-sided limit exists and is also $5$ . Since the limit equals the point value, $f$ is continuous at $x=2$ .

A Derivative Is a Limit of Secant Slopes

At a point $x$ , choose a second input $x+h$ with $h\ne0$ . The corresponding secant slope is the difference quotient:

\frac{f(x+h)-f(x)}{h}.

As $h$ approaches $0$ , the second point approaches the first. If the secant slopes approach a finite value, that value is the derivative:

f'(x) =\lim_{h\to0}\frac{f(x+h)-f(x)}{h}.

For $f(x)=x^2$ :

\begin{aligned} \frac{f(x+h)-f(x)}{h} &=\frac{(x+h)^2-x^2}{h}\\ &=\frac{2xh+h^2}{h}\\ &=2x+h, \qquad h\ne0. \end{aligned}

Taking the limit gives:

f'(x)=\lim_{h\to0}(2x+h)=2x.

The quotient never substitutes $h=0$ before cancellation; doing so would divide by zero. The limit asks which value the valid nonzero-step slopes approach.

Exercise: Simplify a difference quotient

For $f(x)=3x+1$ , what does $[f(x+h)-f(x)]/h$ simplify to when $h\ne0$ ?

Check the simplified quotient

Compute it first, then check your number.

HintExpand before dividing

Compute $3(x+h)+1-(3x+1)$ .

SolutionThe secant slopes are already constant

Substitution gives:

\frac{3(x+h)+1-(3x+1)}{h} =\frac{3h}{h} =3, \qquad h\ne0.

Every valid secant slope of this line is $3$ , so its limiting local slope is also $3$ .

Before Taking a Limit

Identify the input being approached and the proposed output value. Try direct substitution when continuity justifies it; if substitution is indeterminate, study the nearby expression instead. For a two-sided limit, check both directions. Keep the limit separate from the function value until continuity has been established.

The next lesson studies derivatives and local sensitivity. It uses the limit of the difference quotient to calculate local slopes and to build linear approximations near a chosen input.