Limits of Composite Functions

The limit of a composite function f(g(x)) as x approaches x₀ is equal to l:
$$ \lim_{x \rightarrow x_0} f(g(x)) = l $$ provided that:
$$ \lim_{x \rightarrow x_0} g(x) = y_0 $$ and:
$$ \lim_{y \rightarrow y_0} f(y) = l $$ and there exists some δ > 0 such that:
$$ g(x) \ne \lim_{x \rightarrow x_0} g(x) \quad \forall x \ne x_0 \text{ in } (x_0 - \delta, x_0 + \delta) $$

If the function g(x) tends toward y as x → x₀:

$$ \lim_{x \rightarrow x_0} g(x) = y_0 $$

and the function f(y) tends toward l as y → y₀:

$$ \lim_{y \rightarrow y_0} f(y) = l $$

then the composite function f(g(x)) also tends toward l as x → x₀:

$$ \lim_{x \rightarrow x_0} f(g(x)) = l $$

A Practical Example

Consider the function f(x):

$$ f(x) = \log \frac{1}{x} $$

This is a composite function of the form f(g(x)), where:

$$ y = g(x) = \frac{1}{x} $$

and:

$$ f(y) = \log y $$

To compute the limit of the composite function:

$$ \lim_{x \rightarrow \infty} \log \frac{1}{x} $$

we can split the calculation into two separate limits.

First, determine the limit of g(x) as x approaches infinity:

$$ y_0 = \lim_{x \rightarrow \infty} g(x) = \lim_{x \rightarrow \infty} \frac{1}{x} = 0 $$

Then, evaluate the limit of f(y) as y approaches y₀ = 0:

$$ l = \lim_{y \rightarrow 0} f(y) = \lim_{y \rightarrow 0} \log y = -\infty $$

Therefore, the limit of the composite function as x approaches infinity is l = -∞.

Proof

Let’s consider the composite function f(g(x)) converging to l as x approaches x₀:

$$ \lim_{x \rightarrow x_0} f(g(x)) = l $$

Take a sequence x_n within the domain of f(g(x)) such that x_n converges to x₀:

$$ \lim_{n \rightarrow \infty} x_n = x_0 $$

By the definition of the limit of a sequence, the limit as x → x₀ equals l if there exists an index v such that:

$$ |x_n - x_0| < \delta \quad \forall n > v $$

Thus, the sequence x_n excludes the term x₀.

From this sequence x_n, we construct a new sequence y_n via the function g(x):

$$ y_n = g(x_n) $$

Since the sequence does not contain x₀, for all n > v, it follows that y_n = g(x_n) is distinct from g(x₀):

$$ y_n \ne g(x_0) $$

Therefore, since y₀ = g(x₀):

$$ y_n \ne y_0 \quad \forall n > v $$

Moreover, the limit of the sequence y_n converges to l:

$$ \lim_{n \rightarrow \infty} y_n = l $$

And so on.