Is $g\circ f$ continuous at $x = 4$?
$$f(x)=\begin{cases} -1 &, x=4 \\ 1 &, x \ne 4\end{cases}$$
$$g(x)=\begin{cases} -6 &, x=4 \\ 4x-10 & x \ne 4\end{cases}$$
My answer for this is 'Yes' because $g(f(x)) = g(-1) = -6$ and $g(f(x))$ approaches $-6$ as $x$ approaches $4$. Because the limit and $g(f(x))$ are the same, I concluded that the composition is continuous at $x = 4$.
However, the answer in my textbook is 'No'.
Can you help me on pointing out the deficiency in my solution?
$$f(x)=\begin{cases} -1 &, x=4 \\ 1 &, x \ne 4\end{cases}$$
$$g(x)=\begin{cases} -6 &, x=4 \\ 4x-10 & x \ne 4\end{cases}$$
$$g(f(4))=g(-1)=4(-1)-10=\color{red}{-14}$$
If $x \ne 4$,
$$g(f(x))=4(1)-10=-6 \ne -14$$