03 Continuity

Testing Continuity at a Point

Continuity describes whether a function changes without a break at a point or throughout an interval. The graph-based intuition is that a continuous graph can be traced near the point without lifting a pencil, but the precise test uses the function value and a limit.

A function $f$ is continuous at $x=a$ exactly when all three requirements hold:

1. $f(a)$ is defined.

2. $\lim_{x\to a}f(x)$ exists.

3. $\lim_{x\to a}f(x)=f(a)$.

Equivalently, when $f(a)$ is defined,

For example, let $f(x)=x^2+3x-1$. Since every polynomial is defined and continuous for every real number, $f$ is continuous at $x=2$. Directly,

Continuity can fail in three basic ways: the function value may be missing, the limit may fail to exist, or the limit may exist but disagree with the function value.

Takeaway: Always check the value, the limit, and their equality; none of these checks can be omitted.

Continuity on Intervals and Domains

On an open interval $(a,b)$, continuity means that the function is continuous at every point in the interval. On a closed interval $[a,b]$, the interior uses the same pointwise test, while the endpoints use one-sided limits:

A function only needs to be defined on the interval under consideration; it does not need to have a formula outside that interval. For example,

has domain $[-2,2]$ and is continuous on that entire closed interval. At the endpoints,

Common continuity patterns include the following:

• A polynomial is continuous for every real number.

• A rational function is continuous wherever its denominator is nonzero.

• An exponential function is continuous for every real number.

• A logarithmic function is continuous throughout its domain, such as $x>0$ for $\ln x$.

• A root function is continuous wherever the root is real and defined.

• Functions such as $\sin x$ and $\cos x$ are continuous throughout their natural domains.

For $f(x)=\dfrac{x-1}{x^2+2x}$, factor the denominator as $x(x+2)$. The function is therefore continuous on

Takeaway: Determine the domain first, then identify the intervals and included endpoints on which continuity is being claimed.

Classifying Discontinuities

A discontinuity is classified by the behavior of the function and its one-sided limits near the problematic input.

A Removable discontinuity occurs when the finite two-sided limit exists but the function is missing or has the wrong value. For

simplification gives $f(x)=x+3$ for $x\ne 3$, so

The hole can be removed by defining $f(3)=6$.

A Jump discontinuity occurs when the finite one-sided limits are unequal. For

we have

Changing only $f(1)$ cannot make the function continuous because the two sides still approach different values.

An Infinite discontinuity occurs when the function becomes unbounded near a point. For

both one-sided limits equal $+\infty$, so $x=2$ is a vertical asymptote and no finite assigned value can repair the discontinuity.

Some discontinuities arise from oscillation. The function $f(x)=\sin\left(\dfrac{1}{x}\right)$ has no limit as $x\to 0$ because it oscillates increasingly rapidly.

Takeaway: Compare the one-sided limits first. A finite matching limit suggests a hole, unequal finite limits indicate a jump, and unbounded behavior indicates an infinite discontinuity.

Algebraic Operations and Composition

Continuity is preserved by many algebraic operations. If $f$ and $g$ are continuous at $a$, then the following are continuous at $a$:

• $f+g$

• $f-g$

• $cf$, where $c$ is a constant

• $fg$

• $\dfrac{f}{g}$, provided that $g(a)\ne 0$

These rules explain why polynomials are continuous everywhere and why rational functions are continuous at every point in their domains. For example,

is continuous for every real number. The numerator is continuous, and the denominator satisfies

for every real $x$, so the denominator never vanishes.

Composition requires an additional domain check. If $g$ is continuous at $a$ and $f$ is continuous at $g(a)$, then the composite function is continuous at $a$:

For $h(x)=\sqrt{1+x^2}$, the inner function $g(x)=1+x^2$ is continuous everywhere and always satisfies $g(x)\ge 1$, which lies in the domain of the outer function $f(u)=\sqrt{u}$. Thus, $h$ is continuous for every real $x$.

Takeaway: For a composition, check the inner function's continuity and ensure that its outputs stay in the outer function's domain.

Theorems Powered by Continuity

Continuity supports two central existence theorems.

The Extreme Value Theorem says that a function continuous on a closed, bounded interval $[a,b]$ attains both an absolute maximum and an absolute minimum there. The closed interval and continuity hypotheses matter: on an open interval or across a discontinuity, extrema need not be attained.

The Intermediate Value Theorem states that if $f$ is continuous on $[a,b]$ and $N$ lies between $f(a)$ and $f(b)$, then at least one $c\in[a,b]$ satisfies

When the endpoint values have opposite signs,

the theorem guarantees a root in $(a,b)$.

To show that $x^3+x-1=0$ has a solution between $0$ and $1$, define $f(x)=x^3+x-1$. This polynomial is continuous on $[0,1]$, and

Because zero lies between the endpoint values, the Intermediate Value Theorem guarantees some $c\in(0,1)$ with $f(c)=0$. It does not determine the exact root or prove that the root is unique.

A common incorrect application uses $f(x)=1/x$ on $[-1,1]$. Although the endpoint values have opposite signs, the function is not continuous on the whole interval because it is undefined at $x=0$. Therefore, the theorem cannot be applied.

Takeaway: Before using an existence theorem, verify every hypothesis, especially continuity on the entire closed interval.

Inverse Functions and Continuity

An inverse function reverses the input-output roles of a one-to-one function. A function is one-to-one when different inputs produce different outputs; equivalently, every horizontal line intersects its graph at most once.

If $f$ is continuous and one-to-one on an interval $I$, then its inverse $f^{-1}$ is continuous on the range $f(I)$. The inverse has domain equal to the original range and range equal to the original domain.

For $f(x)=3x-5$, solve $y=3x-5$ for $x$:

Therefore,

Because the original function is continuous and one-to-one on $\mathbb{R}$, its inverse is continuous on $\mathbb{R}$.

The function $f(x)=x^2$ is not one-to-one on $\mathbb{R}$, since $f(-2)=f(2)=4$. Restricting its domain to $[0,\infty)$ makes it one-to-one, and the inverse becomes

The identities

express that the two functions undo one another on their relevant domains.

Takeaway: To obtain a continuous inverse, first ensure that the original function is continuous and one-to-one on an interval.

A Practical Continuity Checklist

Use the following procedure when analyzing continuity at $x=a$:

1. Check whether $f(a)$ is defined.

2. Determine $\lim_{x\to a}f(x)$, if it exists.

3. Compare the limit with $f(a)$.

4. If the limit does not exist, compare the one-sided limits.

5. Classify the failure as removable, a jump, infinite, or oscillatory when appropriate.

For continuity on an interval:

1. Determine the function's domain.

2. Locate excluded inputs, such as zeros of a denominator or invalid inputs to an even root or logarithm.

3. Divide the domain into intervals separated by excluded values.

4. Check one-sided continuity at included endpoints.

5. Confirm the hypotheses before applying the Extreme Value Theorem or the Intermediate Value Theorem.

Keep the logical relationships in mind:

• Differentiability at a point implies continuity at that point.

• Continuity does not imply differentiability; $f(x)=|x|$ is continuous but has a corner at $x=0$.

• Continuity is preserved by sums, differences, products, and valid quotients.

• Continuity of the inner and outer functions, together with the composition domain, gives continuity of a composition.

• A continuous one-to-one function on an interval has a continuous inverse on its range.

Final takeaway: Continuity is both a local test involving limits and a global condition that enables powerful conclusions about values, extrema, compositions, and inverses.