Problem 70 - Calculus: Early Transcendental Functions

Question

Let \(f\) be a continuous function on the interval \([0,1],\) and \(\mathrm{de}-\) fine \(g_{n}(x)=f\left(x^{n}\right)\) for \(n=1,2\) and so on. For a given \(x\) with \(0 \leq x \leq 1,\) find \(\lim _{n \rightarrow \infty} g_{n}(x) .\) Then, find \(\lim _{n \rightarrow \infty} \int_{0}^{1} g_{n}(x) d x\)

Step-by-Step Solution

Answer

The limit of \(g_n(x)\) as \(n\) tends to infinity for \(0 \leq x \leq 1\) is \(f(0)\), and the limit of the integral from 0 to 1 of \(g_n(x)dx\) as \(n\) tends to infinity is also \(f(0)\).

Step 1: Evaluate limit of function

In the given exercise, function \(f\) is continuous on the interval [0,1]. The limit of \(g_n(x)\), is needed when \(n\) reaches infinity. For \(0 < x < 1\), \(x^n\) tends to 0 as \(n\) approaches infinity, i.e. \(\lim_{n \rightarrow \infty} x^n = 0\). Given \(g_n(x) = f(x^n)\), and \(f\) is continuous at 0, this results in \(\lim _{n \rightarrow \infty} g_{n}(x) = f(\lim_{n \rightarrow \infty} x^n) = f(0)\). And for \(x = 0\) and \(x = 1\), obviously \(\lim _{n \rightarrow \infty} g_{n}(x) = f(x)\). So, for \(0 \leq x \leq 1\), \(\lim _{n \rightarrow \infty} g_{n}(x) = f(0)\).

Step 2: Evaluate limit of integration

To evaluate \(\lim_{n\rightarrow\infty}\int_0^1g_n(x)dx\), substitute \(g_n(x) = f(x^n)\). Then we get \(\int_0^1 f(x^n) dx\). As the limit of \(g_n(x)\) when \(n\) tends to infinity is \(f(0)\), we get \(\lim_{n\rightarrow\infty} f(x^n) = f(0)\), so \(\lim_{n\rightarrow\infty}\int_0^1g_n(x)dx = \int_0^1 f(0) dx = f(0)\).

Step 3: Conclude the results

Based on the calculations, for a continuous function \(f\) on the interval [0,1] and \(g_n(x) = f(x^n)\), \(\lim_{n\rightarrow\infty}g_n(x) = f(0)\) for \(0 \leq x \leq 1\), and \(\lim_{n\rightarrow\infty}\int_0^1g_n(x)dx = f(0) .\)

Key Concepts

Continuity

Continuity is a fundamental concept in calculus and analysis. When we say that a function, such as \(f\) in this exercise, is continuous on an interval, it means that there are no breaks, jumps, or holes in the graph of the function over that interval. In simpler terms, if you can draw the graph of the function without lifting your pencil, the function is continuous.

For a function \(f\) to be continuous at a point \(c\), the following must be true:

• \(f(c)\) is defined.
• The limit of \(f(x)\) as \(x\) approaches \(c\) exists.
• The limit of \(f(x)\) as \(x\) approaches \(c\) is equal to \(f(c)\).

In the given problem, we assume \(f\) is continuous on the interval \([0,1]\). This allows us to use the property of continuity that extends limits through the function, which is why we can replace \(x^n\) in \(f(x^n)\) with the limit value that \(x^n\) approaches. Thus, as \(x^n\) tends to zero for \(0 < x < 1\) when \(n\) goes to infinity, \(f(x^n)\) tends to \(f(0)\) because \(f\) is continuous.

Definite Integration

Definite integration is the process of calculating the area under a curve bounded by a specified interval, expressed usually as \(\int_{a}^{b} f(x) \, dx \\).

In our problem, we need to evaluate the limit of the integral as \(n\) approaches infinity for the sequence of functions \(g_n(x) = f(x^n)\) between 0 and 1.

Intuitively, the integral \(\int_0^1 f(x^n) \, dx \\) gives the area under \(f(x^n)\) from 0 to 1. As \(n\) grows, \(x^n\) essentially "squeezes" the interval into the point 0 for \(0 < x < 1\), meaning that the limit of the integral becomes \(\int_0^1 f(0) \, dx\), since \(f(x^n)\) converges to \(f(0)\). This results in \(f(0)\) multiplied by 1, because the length of the interval (from 0 to 1) defines the width of a rectangle with constant height \(f(0)\). Therefore, the solution correctly concludes that the integral's limit is \(f(0)\).

Sequence of Functions

A sequence of functions involves a list of functions \(f_1, f_2, f_3, \ldots\) indexed by a natural number, often represented as \(f_n\). These sequences can converge to a function pointwise or uniformly, offering various properties and implications in analysis.

For this exercise, \(g_n(x) = f(x^n)\) describes a sequence of functions where each function depends on \(f\) and varies with \(n\). The sequence \(g_n(x)\) converges pointwise on the interval \([0, 1]\) to \(f(0)\). This is because, for \(0 < x < 1\), \(x^n\), becomes smaller and approaches zero with increasing \(n\), so each \(g_n(x)\) converges to \(f(0)\). At the endpoints, \(x=0\) and \(x=1\), the function values do not change and the convergence remains consistent.

Understanding function sequences and their convergence is essential when determining the behavior of integrals and other calculations as parameters approach limits. Here, the uniform behavior aids in simplifying the integral of \(g_n(x)\) to \(f(0)\), as explored above.