Limit of a Composite Function
Let f(g(x)) be a composite function. We say that $$ \lim_{x \rightarrow x_0} f(g(x)) = l $$ provided that $$ \lim_{x \rightarrow x_0} g(x) = y_0 $$ and $$ \lim_{y \rightarrow y_0} f(y) = l $$ and that there exists a δ>0 such that $$ g(x) \ne y_0 \:\:\: \forall x \ne x_0 \in (x_0-δ, x_0+δ) $$ In other words, the limit of the composite function equals l whenever the inner function approaches y0, the outer function approaches l as its argument approaches y0, and g(x) does not take the value y0 in a deleted neighborhood of x0.
To evaluate the limit of a composite function as x→x0 , proceed in two stages:
$$ \lim_{x \to x_0} f(g(x)) $$
First, compute the limit of the inner function:
$$ \lim_{x \rightarrow x_0} g(x) = y_0 $$
Then compute the limit of the outer function at y0:
$$ \lim_{y \rightarrow y_0} f(y) = l $$
If this second limit exists, it follows that
$$ \lim_{x \rightarrow x_0} f(g(x)) = l $$
Special case. If the inner function \( g(x) \) is continuous at \( x_0 \) and the outer function \( f(z) \) is continuous at \( z_0 = g(x_0) \), then the composite function \( f(g(x)) \) is continuous at \( x_0 \). In this setting, the limit is obtained by direct substitution: \[ \lim_{x \to x_0} f(g(x)) = f(g(x_0)) \] This is the standard result stating that continuity is preserved under composition.
Example 1
Consider the function
$$ f(x) = \log \frac{1}{x} $$
This can be expressed as a composite function f(g(x)), where
$$ y = g(x) = \frac{1}{x} $$
$$ f(y) = \log y $$
We compute
$$ \lim_{x \rightarrow \infty} \log \frac{1}{x} $$
First evaluate the inner limit:
$$ y_0 = \lim_{x \rightarrow \infty} g(x) = \lim_{x \rightarrow \infty} \frac{1}{x} = 0 $$
Next evaluate the outer limit. Since the logarithm is defined only for positive arguments, we consider the right-hand limit:
$$ l = \lim_{y \rightarrow 0^+} \log y = -\infty $$
Therefore,
$$ \lim_{x \rightarrow \infty} \log \frac{1}{x} = -\infty $$
Example 2
Consider the composite function
$$ f(g(x)) = \sin(4x) $$
The inner function is
$$ z = g(x) = 4x $$
and the outer function is
$$ f(z) = \sin z $$
Both functions are continuous on \( \mathbb{R} \).
Compute
$$ \lim_{x \to \frac{\pi}{4}} \sin(4x) $$
First evaluate the inner limit:
$$ \lim_{x \to \frac{\pi}{4}} 4x = 4 \cdot \frac{\pi}{4} = \pi $$
Since the sine function is continuous at \( \pi \), we substitute directly:
$$ \sin(\pi) = 0 $$
Hence,
$$ \lim_{x \to \frac{\pi}{4}} \sin(4x) = 0 $$
Example 3
Compute
$$ \lim_{x \to 2} \sqrt{x^2 + 1} $$
The inner function is
$$ z = g(x) = x^2 + 1 $$
The outer function is
$$ f(z) = \sqrt{z} $$
First evaluate the inner limit:
$$ \lim_{x \to 2} (x^2 + 1) = 4 + 1 = 5 $$
Since the square root function is continuous at \( z = 5 \), we substitute directly:
$$ f(5) = \sqrt{5} $$
Therefore,
$$ \lim_{x \to 2} \sqrt{x^2 + 1} = \sqrt{5} $$
Proof
Assume that
$$ \lim_{x \rightarrow x_0} f(g(x)) = l $$
Let \( \{x_n\} \) be a sequence in the domain of f(g(x)) such that
$$ \lim_{n \rightarrow \infty} x_n = x_0 $$
By the sequential characterization of limits, for every δ>0 there exists an index v such that
$$ |x_n - x_0| < \delta \:\:\: \forall n > v $$
Define the sequence
$$ y_n = g(x_n) $$
Since
$$ \lim_{x \rightarrow x_0} g(x) = y_0 $$
it follows that
$$ \lim_{n \rightarrow \infty} y_n = y_0 $$
Moreover, because g(x) does not take the value y0 in a deleted neighborhood of x0, we have
$$ y_n \ne y_0 \:\:\: \forall n > v $$
Using the sequential characterization once more, the assumption
$$ \lim_{y \rightarrow y_0} f(y) = l $$
implies
$$ \lim_{n \rightarrow \infty} f(y_n) = l $$
Therefore,
$$ \lim_{n \rightarrow \infty} f(g(x_n)) = l $$
This completes the proof.
