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Math Analysis 1

Limit of a composite function: theorem, proof and change of variable

When you can move a limit inside a function: statement, epsilon-delta proof, a counterexample, change of variable and worked examples.

Updated 19 October 2026 3 min read 0 comments

When you compute a limit such as \lim_{x\to 0} \sqrt{1+\sin x} it feels natural to “move the limit inside”: first find the limit of 1 + \sin x, which is 1, then take the square root. You do this dozens of times in every exercise, but it is not always allowed. The theorem on the limit of a composite function says exactly when it is.

The statement

Let g be defined in a neighbourhood of c (except possibly at c) and f defined in a neighbourhood of L. If

  • \displaystyle\lim_{x \to c} g(x) = L, and
  • f is continuous at L,

then

    \[ \lim_{x \to c} f\big(g(x)\big) = f(L) = f\Big(\lim_{x \to c} g(x)\Big). \]

In words: a continuous function “commutes” with the limit.

The epsilon-delta proof

Fix \varepsilon > 0. We need a \delta > 0 such that 0 < |x - c| < \delta implies |f(g(x)) - f(L)| < \varepsilon. We build it in two steps, starting from the end.

  1. Continuity of f at L. There is \eta > 0 such that |y - L| < \eta \Rightarrow |f(y) - f(L)| < \varepsilon. So it is enough for g(x) to land within \eta of L.
  2. Limit of g. Apply the definition of limit with \eta in place of epsilon: there is \delta > 0 such that 0 < |x - c| < \delta \Rightarrow |g(x) - L| < \eta.

Putting the two together: if 0 < |x - c| < \delta, then |g(x) - L| < \eta, and therefore |f(g(x)) - f(L)| < \varepsilon. Since \varepsilon was arbitrary, the limit is f(L). ∎

The key point is that the \delta for g is chosen after receiving the \eta from f: the two definitions mesh like two gears.

Why continuity is needed: a counterexample

If f only has a limit at L, without being continuous there, the result can fail. Take

    \[ g(x) = 0 \ \text{for every } x, \qquad f(y) = \begin{cases} 1 & y = 0 \\ 0 & y \neq 0 \end{cases} \]

We have \lim_{x\to 0} g(x) = 0 and \lim_{y \to 0} f(y) = 0. But f(g(x)) = f(0) = 1 for every x, so \lim_{x \to 0} f(g(x)) = 1, not 0. The trouble is that g “lands” exactly on the point where f jumps.

There is also a version of the theorem without continuity: if \lim_{y\to L} f(y) = M and g(x) \neq L in a neighbourhood of c (except at c), then \lim_{x\to c} f(g(x)) = M. This is what justifies a change of variable in limits.

How it is used in exercises

Example 1. \displaystyle \lim_{x \to 0} e^{\frac{\sin x}{x}}. The exponent tends to 1 (a standard limit) and the exponential is continuous, so the limit is e^1 = e.

Example 2 (change of variable). \displaystyle \lim_{x \to 0} \frac{\sin(3x)}{3x}. Set t = 3x: as x \to 0, also t \to 0, and t \neq 0 when x \neq 0. The limit becomes \lim_{t\to 0}\frac{\sin t}{t} = 1.

Example 3. \displaystyle \lim_{x \to +\infty} \ln\!\left(\frac{2x+1}{x}\right). The argument tends to 2, the logarithm is continuous at 2, so the limit is \ln 2.

Almost every function you meet (polynomials, exponentials, logarithms, sine and cosine, roots) is continuous on its domain, which is why “moving the limit inside” works so often. Mistakes happen when the argument tends to a point outside the domain (the logarithm of something that tends to 0) or to a point of discontinuity, such as the integer part of a number that tends to an integer.

The same theorem with sequences

Continuity can also be characterised with sequences: f is continuous at L if and only if f(y_n) \to f(L) for every sequence y_n \to L. This gives the same result: if a_n \to L and f is continuous at L, then f(a_n) \to f(L). That is why, for instance, \sqrt{1 + 1/n} \to 1.

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