Continuity composite functions

Suppose that $f:D\rightarrow(-\infty,\infty)$,$g:E\rightarrow(-\infty,\infty)$, and the range of f is contained in E. Then it makes sense to define $g\circ f:D\rightarrow(-\infty,\infty)$ by the rule $(g\circ f)(x) = g(f(x))$. What we will now show is that if f is continous at a and g is continuous at f(a) then $g\circ f$ is continuous at a. To do this we have to show that

$$\lim_{x\rightarrow a}(g\circ f)(x) = g(f(a))$$

That is, for every t > 0 we have to produce m_t so that if $x\in D$ and 0 < |x-a| < m_t then $\vert(g\circ f)(x) - g(f(a))\vert < t$. We will go in two stages. Remember first that g is continuous at f(a), so that for every t > 0 we can find p_t > 0 so that if $u\in E$ and 0 < |u - f(a)| < p_t then |g(u) - g(f(a))| < t. Secondly, remember that since f is continuous at a, given this p_t > 0 we can find m_t so that if $x\in D$ and 0 < |x-a| < m_t then |f(x) - f(a)| < p_t.

Now we reason as follows. If $x\in D$ and 0 < |x-a| < m_t then |f(x) - f(a)| < p_t. If f(x) = f(a) then |g(f(x))-g(f(a))| = 0 < t. If $f(x) \neq f(a)$ then $f(x)\in E$ and 0 < |f(x) - f(a)| < p_t, so |g(f(x)) - g(f(a))| < t. Since $\vert g(f(x)) - g(f(a))\vert = \vert(g\circ f)(x) - g(f(a))\vert$ we have shown that if $x\in D$ and 0 < |x-a| < m_t then $\vert(g\circ f)(x) - g(f(a))\vert < t$, as required.