A function \(f(x)\) is continuous at a point \(x=a\) if and only if the following conditions are satisfied:
- \(f(a)\) exists (i.e., \(a\) lies in the domain of \(f\).
- \(\lim_{x\to a}f(x)\) exists (\(f\) has a limit as \(\mathcal{X}\to\mathcal{a}\), i.e. the limit is a real number)
- \(\lim_{x\to a}f(x) = f(a)\) (the limit equals the function value)
If one or more of these conditions fail, then \(f(x)\) is not continuous at a point \(x=a\).
Example: Consider the function \(\left.f(x)=\left\{\begin{aligned}\frac{x^3+2x^2-8x}{x^2-4}&&,&&\mathrm{if~}x\neq2\\4&&,&&\mathrm{if~}x=2\end{aligned}\right.\right..\)
- Is \(f(x)\) continuous at \(x=-2\)?
- Does \(\lim_{x\to 2}f(x)\) exist?
- Is \(f(x)\) continuous at \(x=2\)?
Solution:
- \(f(-2)=\frac{\left(-2\right)^3+2(-2)^2-8(-2)}{\left(-2\right)^2-4}=\frac{16}0.\) Thus \(f(-2)\) is not defined, and hence \(f(x)\) is NOT continuous at \(x=-2\).
- Note that one-side is not necessary, but for completeness, it is shown.
\(\begin{aligned}
\operatorname*{lim}_{x\rightarrow2^{-}}f(x)& \begin{aligned}&=\lim_{x\to2^-}\frac{x^3+2x^2-8x}{x^2-4}\end{aligned} \\
&=\lim_{x\to2^-}\frac{x(x+4)(x-2)}{(x+2)(x-2)}=\lim_{x\to2^-}\frac{x(x+4)}{x+2} \\
&=\frac{2(2+4)}{2+2}=3
\end{aligned}\)
\(\lim_{x\to2^+}f(x)=\lim_{x\to2^+}\frac{x(x+4)}{x+2}=\frac{2(2+4)}{2+2}=3\)
Yes, the limit exists, that is \(\lim_{x\to 2}f(x)=3\).
- Note that \(f(2)=4\neq3=\lim_{x\to2}f(x)\). By the definition of continuity, \(f(x)\) is NOT continuous at \(x=2\).