# Suppose that $f(x) \to\ell$ as $x\to a$ and $g(y) \to k$ as $y \to\ell$. Does it follow that $g(f(x)) \to k$ as $x \to a$?

If $$f$$ and $$g$$ are continuous the limit of the composition is the composition of the limit, so the implication follows. But what if $$f$$ and $$g$$ are not continuous? There exist counter examples to the implication? Or does it follow also in this case?

No, not necessarily: we can abuse the fact that the definition of limit ignores the value the function may take there.

For example, take $$a = l = k = 0$$ and let $$f,g:\mathbb R \to \mathbb R$$ be as follows.

$$f(x) := 0,\qquad\text{ and}$$

$$g(x) := \begin{cases} 1 & x = 0 \\ 0 & x \neq 0\end{cases}.$$

Then:

$$\lim_{x\to 0} g(x) = \lim_{x\to 0} f(x) = 0,$$

but $$g(f(x)) = 1$$ for all $$x \in \mathbb R$$, so $$\displaystyle\lim_{x \to 0} g(f(x)) = 1$$, not $$0$$.

Claim The statement in the title holds if and only if one of the following hold:

• $$g$$ is continuous at $$l$$, or
• There is some neighbourhood of $$a$$ (excluding $$a$$) on which $$f$$ does not obtain the value $$l$$.

Proof

Case 1. Assume $$g$$ is continuous at $$l$$.

Let $$\epsilon > 0$$. Since $$\lim_{y \to l}g(y) = k$$ and $$g(y) = k$$ (by continuity), there exists $$\delta>0$$ such that

$$0 \boldsymbol{\leq} |y - l| < \delta \implies |g(y)-k| < \epsilon. \tag{1}$$

(Note that we can only use "$$\leq$$" at $$\ast$$ because it is continuous!)

Since $$\lim_{x \to a}f(x) = l$$, there exists $$\eta>0$$ such that

$$0 < |y - l| < \eta \implies |f(x)-l| < \delta. \tag{2}$$

which, combined with one, gives you the limit you want.

Case 2. Assume there is some neighbourhood of $$a$$ (excluding $$a$$) on which $$f$$ does not obtain $$l$$ at all: i.e., there exists $$\tau>0$$ such that

$$0<|x - a|<\tau \implies f(x) \neq l. \tag{3}$$

Then, repeating the argument for the first case:

Given $$\epsilon > 0$$, since $$\lim_{y \to l}g(y) = k$$, there exists $$\delta>0$$ such that

$$0 < |y - l| < \delta \implies |g(y)-k| < \epsilon. \tag{4}$$

(And without continuity, that's all we get.)

Since $$\lim_{x \to a}f(x) = l$$, there exists $$\eta>0$$ such that

$$0 < |y - l| < \eta \implies |f(x)-l| < \delta. \tag{5}.$$

But moreover, combining $$(5)$$ with $$(3)$$ gives

$$0 < |y - l| < \min(\eta,\tau) \implies 0 < |f(x)-l| < \delta, \tag{6}$$

which you can combine with $$(4)$$ to get the desired limit.

Case 3. Assume $$g$$ is discontinuous at $$l$$, and that $$f$$ attains the value $$l$$ on any (arbitrarily small) neighbourhood of $$a$$.

We'll use a sequence. You can prove (by contradiction — with the second assumption) that there must be a sequence of points $$(x_n)_{n=1}^∞ \to a,$$ with $$x_n \neq a$$ for any $$n$$, such that, for every $$n \in \mathbb N$$,

$$f(x_n) = l.$$

Consequently,

$$g(f(x_n)) = g(l),$$

and by the discontinuity of $$g$$ at $$l$$, $$k \neq g(l)$$.

For contradiction, assume that $$fg(x) \to k$$ as $$x \to a$$. Then (by the equivalent sequential definition of continuous limit), every sequence $$(k_n)_{n=1}^\infty\to a$$ with $$k_n \neq a$$ satisfies

$$g(f(k_n)) \to k.$$

But aha! $$(x_n)_n$$ is exactly an example of a sequence for which $$g(f(k_n)) \not\to k$$ (because it already converges to $$g(l)\neq k$$).

We can construct some simple functions as counter-example: Define $$f:A \to B$$ and $$g:B \to C$$ as $$f(x) = y_0 \quad \text{(constant function)}$$ and $$g(y) = z_1 \text{ if }y \neq y_0 \text{ and } g(y) = z_2 \text{ if }y = y_0.$$ Then $$f(x) \to y_0$$ as $$x \to x_0$$ (arbitrary point), and $$g(y) \to z_1$$ as $$y \to y_0$$. But $$g(f(x)) \to z_2$$ as $$x \to x_0$$.