University Calculus: Early Transcendentals (3rd Edition)

We can extend a function $y=f(x)$ to be continuous at a point $x=c$ if $f(x)$ has a removable discontinuity at $x=c$.
We can extend a function $y=f(x)$ to be continuous at a point $x=c$ if $f(x)$ has a removable discontinuity at $x=c$, which means: - Both $\lim_{x\to c}f(x)$ and $f(c)$ exist. - But $\lim_{x\to c}f(x)\ne f(c)$ In this case, we can extend $f(x)$ to cover a value of $f(c)=\lim_{x\to c}f(x)$ so that $y=f(x)$ will be continuous at $x=c$. For example, take into consideration this function: $f(x)=4x$ for $x\ne 7$ and $f(x)=x$ for $x=7$: - $\lim_{x\to7}f(x)=4\times7=28$ and $f(7)=7$ - As $\lim_{x\to7}f(x)\ne f(7)$, $f(x)$ is discontinuous at $x=7$. But since both $\lim_{x\to7}f(x)$ and $f(7)$ still exist, this discontinuity is removable. - That means we can extend $f(x)$ to include a value of $f(7)=28$ so that $\lim_{x\to7}f(x)=f(7)$, thus making $f(x)$ continuous at $x=7$.