Limit of Composite Function

Theorem
Let $$f$$ and $$g$$ be real functions.

Let:
 * $$\lim_{y \to \eta} f \left({y}\right) = l$$;
 * $$\lim_{x \to \xi} g \left({x}\right) = \eta$$.

Then, if either:
 * Hypothesis 1: $$f$$ is continuous at $\eta$ (i.e. $$l = f \left({\eta}\right)$$, or;
 * Hypothesis 2: for some open interval $$I$$ containing $$\xi$$, it is true that $$g \left({x}\right) \ne \xi$$ for any $$x \in I$$ except possibly $$x = \xi$$,

then $$\lim_{x \to \xi} f \left({g \left({x}\right)}\right) = l$$.

Continuity
Let $$I$$ and $$J$$ be real interval.

Let:
 * $$g: I \to J$$ be a real function which is continuous on $$I$$;
 * $$f: J \to \mathbb{R}$$ be a real function which is continuous on $$J$$.

Then the composite function $$f \circ g$$ is continuous on $$I$$.

Proof
Let $$\epsilon > 0$$.

Since $$\lim_{y \to \eta} f \left({y}\right) = l$$, we can find $$\Delta > 0$$ such that $$\left|{f \left({y}\right) - l}\right| < \epsilon$$ provided $$0 < \left|{y - \eta}\right| < \Delta$$.

Let $$y = g \left({x}\right)$$.

Then, provided that $$0 < \left|{g \left({x}\right) - \eta}\right| < \Delta$$, we have $$\left|{f \left({g \left({x}\right)}\right) - l}\right| < \epsilon$$.

But $$\lim_{x \to \xi} g \left({x}\right) = \eta$$ and $$\Delta > 0$$.

Hence $$\exists \delta > 0: \left|{g \left({x}\right) - \eta}\right| < \Delta$$ provided that $$0 < \left|{x - \xi}\right| < \delta$$.

We now need to establish the reason for the conditions under which $$0 < \left|{x - \xi}\right| < \delta \Longrightarrow \left|{g \left({x}\right) - \eta}\right| < \Delta$$.

As it stands, this is not generally the case, as follows.

Consider the functions: $$g \left({x}\right) = \eta, f \left({y}\right) = \begin{cases} y_1 & : y = \eta \\ y_2 & : y \ne \eta \end{cases}$$

Then $$\lim_{y \to \eta} f \left({y}\right) = y_2$$ and $$\lim_{x \to \xi} g \left({x}\right) = \eta$$.

But it is not true that $$\lim_{x \to \xi} f \left({g \left({x}\right)}\right) = y_2$$ because $$\forall x: f \left({g \left({x}\right)}\right) = y_1$$.

Now, if Hypothesis 1: $$f$$ is continuous at $\eta$, then $$l = f \left({\eta}\right)$$ and so $$\left|{f \left({y}\right) - l}\right| < \epsilon$$ even when $$y = \eta$$.

So we can write: provided that $$\left|{g \left({x}\right) - \eta}\right| < \Delta$$, we have $$\left|{f \left({g \left({x}\right)}\right) - l}\right| < \epsilon$$, and the argument holds.

Otherwise, let us assume Hypothesis 2: For some open interval $$I$$ containing $$\xi$$, it is true that $$g \left({x}\right) \ne \xi$$ for any $$x \in I$$ except possibly $$x = \xi$$.

Then we can be sure that $$g \left({x}\right) \ne \eta$$ provided that $$0 < \left|{x - \xi}\right| < \delta$$ for sufficiently small $$\delta > 0$$.

But then again we can say that $$\left|{g \left({x}\right) - \eta}\right| < \Delta$$ provided that $$\left|{x - \xi}\right| < \delta$$, and once more the argument holds.

Proof of Continuity
This follows directly and trivially from the definitions of continuity at a point and continuity on an interval.