How to check if a function is continuous by verifying that the function is defined at a point, that its limit exists there, and that the limit equals the function’s actual value. This three-part test is the foundation for analyzing polynomial, rational, radical, trigonometric, and piecewise functions Worth keeping that in mind..
Introduction
Continuity describes whether a function changes without an unexpected break, jump, hole, or vertical asymptote. A graph that can be drawn without lifting a pencil may provide a useful visual clue, but a reliable answer requires checking the function’s limit and function value.
The concept applies both at one input value and across an entire interval. Take this: a function may be continuous everywhere except at one point, or it may be continuous only on a restricted part of its domain. Learning how to check if a function is continuous therefore requires attention to both algebra and domain restrictions But it adds up..
What Continuity at a Point Means
A function (f) is continuous at (x=c) when all three conditions below are satisfied:
- The function is defined at (c): (f(c)) exists.
- The limit exists at (c): (\lim_{x\to c} f(x)) exists.
- The limit equals the function value:
[ \lim_{x\to c} f(x)=f(c). ]
If any condition fails, the function is discontinuous at (c) Practical, not theoretical..
The second condition can also be tested with one-sided limits. The two-sided limit exists only when:
[ \lim_{x\to c^-} f(x)=\lim_{x\to c^+} f(x). ]
The left-hand limit follows inputs below (c), while the right-hand limit follows inputs above (c). If these values differ, the graph has a jump at that point.
Step-by-Step Method for Checking Continuity
1. Identify the point being tested
Begin with the exact input value (x=c). Still, for a piecewise function, pay special attention to every boundary where the formula changes. For a rational function, also identify values that make the denominator zero.
If the task asks whether a function is continuous on an interval, check all relevant points inside that interval. Isolated points outside the interval do not affect continuity on the interval Simple as that..
2. Determine whether (f(c)) exists
Substitute (c) into the function or select the correct piece of a piecewise formula. Check whether the result is a real, defined number Not complicated — just consistent. Less friction, more output..
Common reasons (f(c)) may not exist include:
- Division by zero
- An even root of a negative number
- A logarithm of zero or a negative number
- A missing definition in a piecewise function
- A trigonometric expression outside its domain
If (f(c)) does not exist, the function cannot be continuous at (c), and the remaining conditions do not need to be checked at that point Worth knowing..
3. Calculate the limit as (x) approaches (c)
Evaluate (\lim_{x\to c} f(x)). Direct substitution is often the fastest method when it produces a defined number. If substitution gives an indeterminate form such as (0/0), simplify the expression first Which is the point..
Useful techniques include:
- Factoring and canceling common factors
- Rationalizing a numerator or denominator
- Applying trigonometric identities
- Separating a piecewise function into its left and right formulas
- Comparing the degrees of polynomials in rational expressions
For a piecewise function, calculate both:
[ \lim_{x\to c^-} f(x) \quad\text{and}\quad \lim_{x\to c^+} f(x). ]
The two-sided limit exists only if the one-sided limits are equal and finite.
4. Compare the limit with (f(c))
Once both values are known, place them side by side:
[ L=\lim_{x\to c} f(x), \qquad y=f(c). ]
If (L=y), the function is continuous at (c). So if (L\neq y), the function has a discontinuity even though both values may be defined. This situation often appears as a hole in the graph with a separate point plotted elsewhere.
5. Check every relevant point
A function is continuous on an open interval when it is continuous at every point in that interval. On a closed interval ([a,b]), it must also be continuous from the right at (a) and from the left at (b).
For standard functions, known continuity properties can reduce repeated work:
- Polynomials are continuous for every real number.
- Rational functions are continuous wherever their denominators are nonzero.
- Root functions are continuous throughout their domains.
- Trigonometric functions are continuous throughout their domains.
- Combinations and compositions of continuous functions are continuous where the resulting expressions are defined.
These properties do not remove the need to check boundaries in piecewise functions or excluded values in rational
Applying the Continuity Checklist to Real‑World Functions
The abstract checklist above becomes most useful when you work through a concrete function. Below are three typical scenarios—each highlighting a different source of discontinuity and illustrating how the five‑step method resolves the question.
1. A Piecewise Function with a Potential Jump
Consider
[ f(x)= \begin{cases} x^{2}+1, & x<0,\[4pt] 2x-3, & 0\le x<2,\[4pt] \displaystyle\frac{1}{x-2}, & x\ge 2 . \end{cases} ]
Step 2 – Does (f(0)) exist?
At (x=0) we use the second piece: (f(0)=2(0)-3=-3). The value is defined and real Simple, but easy to overlook..
Step 3 – Limits at the critical points
At (x=0):
[ \lim_{x\to0^-}f(x)=0^{2}+1=1,\qquad \lim_{x\to0^+}f(x)=2(0)-3=-3. ]
Since the one‑sided limits differ, the two‑sided limit does not exist, and the function is discontinuous at (x=0) (a jump discontinuity).
At (x=2):
[ \lim_{x\to2^-}f(x)=2(2)-3=1,\qquad \lim_{x\to2^+}f(x)=\lim_{x\to2^+}\frac{1}{x-2}=+\infty . ]
Again the one‑sided limits are unequal, so (f) is not continuous at (x=2) (a vertical asymptote) The details matter here..
Step 4 – Comparison – Not needed because the limits already fail to exist.
Result: The function is continuous everywhere except at the two breakpoints (x=0) and (x=2).
2. A Rational Function with a Removable Hole
Let
[ g(x)=\frac{x^{2}-4}{x-2}. ]
Step 2 – Is (g(2)) defined?
Direct substitution gives (\frac{0}{0}), an indeterminate form, so (g(2)) is not defined (the original formula excludes (x=2)).
Step 3 – Limit as (x\to2)
Factor the numerator:
[ g(x)=\frac{(x-2)(x+2)}{x-2}=x+2\quad\text{for }x\neq2. ]
Thus
[ \lim_{x\to2}g(x)=2+2=4. ]
Step 4 – Compare – Since (g(2)) does not exist, the function cannot be continuous at (x=2). The graph has a “hole’’ at ((2,4)) It's one of those things that adds up..
Result: (g) is continuous on (\mathbb{R}\setminus{2}); the discontinuity at (x=2) is removable.
3. A Trigonometric Function with Domain Restrictions
Define
[ h(x)=\sqrt{\sin x}, ]
where the square‑root is taken over real numbers only.
Step 2 – Domain of (h)
The radicand (\sin x) must be non‑negative, so (h) is defined only when (\sin x\ge0). This occurs on intervals ([2k\pi,, (2k+1)\pi]) for any integer (k).
Step 3 – Limits at interior points
Inside any such interval, (\sin x) is a continuous function and the square‑root is continuous on ([0,\infty)). By composition, (h) is continuous at every interior point where it is defined It's one of those things that adds up..
Step 4 – Limits at the endpoints
At the left endpoint (x=2k\pi), (\sin x=0) and (h(2k\pi)=0). The right‑hand limit equals the same value, so continuity holds from the right And that's really what it comes down to..
At the right endpoint (x=(2k+1)\pi), (\sin x=0) again,
