When the value at a point agrees with what the nearby values approach.
The idea
In Limits, we separated the value a function approaches from the value it has. Continuity asks them to agree. At a continuous point, nearby inputs give outputs near the actual value there.
A limit alone is not enough. The value at the point must meet the approaching values.
\(x\)\(y\)
Left: \(f(2)=\) Right:
Drag either curve up or down, and move the point at \(x=2\). They turn green when all three heights agree.
Why direct substitution works
Continuity is the reason we can evaluate a limit by substitution. If \(f\) is continuous at \(a\), then
\[\lim_{x\to a}f(x)=f(a).\]
So, for a polynomial, we can substitute the approaching input. For a rational function, we can do this wherever its denominator is nonzero. Substitution is justified by continuity; it is not a rule that applies to every function.
Rules
Building continuous functions
Polynomials are continuous everywhere. Rational functions are continuous wherever their denominator is nonzero. The square-root function is continuous on its domain, with a right-hand limit at its endpoint. Sine and cosine are continuous everywhere.
Sums, differences, and products of continuous functions are continuous. Quotients are continuous where the denominator is nonzero. A composition \(f(g(x))\) is continuous at \(a\) when \(g\) is continuous at \(a\) and \(f\) is continuous at \(g(a)\), using inputs within the relevant domains.
These rules follow from the limit laws; continuity of sine and cosine follows from the trigonometric limits and angle-addition identities. For a piecewise definition, the rules apply within each piece; the joins still need checking.
Checking continuity
When continuity is not already known, find the function value and the limit separately, then compare them. For a piecewise function, check the two sides at each joining point.
Example 1
A value that already fits
For \(p(x)=x^3-2x+1\) at \(x=-1\),
\[p(-1)=2,\qquad \lim_{x\to-1}p(x)=2.\]
Because polynomials are continuous, substitution gives the limit directly: \(\lim_{x\to-1}p(x)=p(-1)=2\). We do not need a separate limit calculation.
