Problem 46 - Calculus Early Transcendentals: Pearson New International Edition

Question

Let \(c_{1}, c_{2}, \ldots, c_{n}\) be positive constants with \(\sum_{i=1}^{n} c_{i}=1\), and let \(x_{1}, x_{2}, \ldots, x_{n}\) be positive numbers. Take natural logarithms and then use l'Hôpital's Rule to show that $$ \lim _{t \rightarrow 0^{+}}\left(\sum_{i=1}^{n} c_{i} x_{i}^{t}\right)^{1 / t}=x_{1}^{c_{1}} x_{2}^{c_{2}} \cdots x_{n}^{c_{n}}=\prod_{i=1}^{n} x_{i}^{c_{i}} $$ Here \(\prod\) means product; that is, \(\prod_{i=1}^{n} a_{i}\) means \(a_{1} \cdot a_{2} \cdots \cdots a_{n} .\) In particular, if \(a, b, x\), and \(y\) are positive and \(a+b=1\), then $$ \lim _{t \rightarrow 0^{+}}\left(a x^{t}+b y^{t}\right)^{1 / t}=x^{a} y^{b} $$

Step-by-Step Solution

Answer

The limit evaluates to \(\prod_{i=1}^{n} x_{i}^{c_{i}}\).

Step 1: Understanding the problem

We need to prove that \(\lim _{t \rightarrow 0^{+}}\left(\sum_{i=1}^{n} c_{i} x_{i}^{t}\right)^{1 / t}=\prod_{i=1}^{n} x_{i}^{c_{i}}\). This includes showing how the weighted sum of powers behaves as \(t\) approaches zero, ultimately resembling a product of powers.

Step 2: Apply Natural Logarithms

Take the natural logarithm of both sides of the expression inside the limit: \[ \ln \left( \left(\sum_{i=1}^{n} c_{i} x_{i}^{t}\right)^{1 / t}\right) = \frac{1}{t} \ln \left(\sum_{i=1}^{n} c_{i} x_{i}^{t}\right). \] This transforms the expression into a form amenable for application of l'Hôpital's rule.

Step 3: Examine the Limit Expression

We now have the limit \[ \lim_{t \to 0^+} \frac{\ln \left(\sum_{i=1}^{n} c_{i} x_{i}^{t}\right)}{t}. \] As \(t\to 0^+\), each \(x_i^t\to 1\), so the numerator and denominator both approach 0, making it suitable for l'Hôpital's rule.

Step 4: Use l'Hôpital's Rule

Differentiate the numerator and the denominator with respect to \(t\): - Derivative of the numerator: \[ \frac{d}{dt} \ln \left(\sum_{i=1}^{n} c_{i} x_{i}^{t}\right) = \frac{\sum_{i=1}^{n} c_{i} x_{i}^{t} \ln x_{i}}{\sum_{i=1}^{n} c_{i} x_{i}^{t}} \]- Derivative of the denominator: \[ \frac{d}{dt}(t) = 1. \]Thus the limit is \[ \lim_{t \to 0^+} \frac{\sum_{i=1}^{n} c_{i} x_{i}^{t} \ln x_{i}}{\sum_{i=1}^{n} c_{i} x_{i}^{t}}. \]

Step 5: Evaluate the Limit

Evaluate the limit expression by substituting \(t \to 0^+\): - Each \(x_i^t\to 1\), so the fraction simplifies to \[ \lim_{t \to 0^+} \sum_{i=1}^{n} c_{i} \ln x_{i} = \sum_{i=1}^{n} c_{i} \ln x_{i}. \]Use properties of logarithms to restructure: \[ \ln \prod_{i=1}^{n} x_{i}^{c_{i}} = \sum_{i=1}^{n} c_{i} \ln x_{i}. \]Thus, \( \ln \prod_{i=1}^{n} x_{i}^{c_{i}} \) equals the limit.

Step 6: Exponentiate to Solve the Limit

Exponentiate both sides to eliminate the logarithm: \[ e^{\ln \prod_{i=1}^{n} x_{i}^{c_{i}}} = \prod_{i=1}^{n} x_{i}^{c_{i}}. \]Hence,\[ \lim _{t \rightarrow 0^{+}}\left(\sum_{i=1}^{n} c_{i} x_{i}^{t}\right)^{1 / t}=\prod_{i=1}^{n} x_{i}^{c_{i}}. \]

Key Concepts

Limit of a Function

In calculus, the "limit of a function" is a fundamental concept that investigates the behavior of a function as the input approaches a particular point. When we say \( \lim_{t \to 0^+} f(t) \), we're interested in understanding what \( f(t) \) approaches as \( t \) gets closer and closer to 0 from the positive side.
The limit does not necessarily need the function to be defined at that specific point; rather, it cares about the value that \( f(t) \) approaches.Understanding limits is crucial because it forms the basis for defining derivatives and integrals, which are core components of calculus. Limits are used to address situations where direct substitutions could lead to seemingly undefined expressions like \( \frac{0}{0} \).
l'Hôpital's Rule helps by providing a method to resolve these indeterminate forms. This rule involves taking derivatives of the numerator and denominator of a function in a limit expression, provided the conditions fit, until the limit is no longer indeterminate.

Natural Logarithm

The natural logarithm, denoted as \( \ln(x) \), is the logarithm to the base \( e \), where \( e \) is an irrational constant approximately equal to 2.71828. It helps transform expressions, especially in contexts where multiplicative relationships can be converted into additive ones. This property is particularly useful when dealing with complex limits and is crucial in the step-by-step solution we are analyzing.
By taking the natural logarithm, we're able to reduce expressions and make them more manageable for analytical techniques like differentiation. For instance, in the given exercise, we transformed \( \left(\sum_{i=1}^{n} c_{i} x_{i}^{t}\right)^{1/t} \) by applying the natural logarithm, resulting in a simpler form \( \frac{1}{t} \ln \left(\sum_{i=1}^{n} c_{i} x_{i}^{t}\right) \) that is suitable for l'Hôpital's Rule. This move simplifies the overall comparison and makes it feasible to tackle the indeterminate form through calculus.

Product of Powers

The product of powers refers to expressions that involve multiplying bases that are raised to exponents. In mathematical notation, if we have numbers \( x_1, x_2, \ldots, x_n \) raised to powers \( c_1, c_2, \ldots, c_n \) respectively, we express their product as \( x_1^{c_1} x_2^{c_2} \cdots x_n^{c_n} = \prod_{i=1}^{n} x_{i}^{c_{i}} \). This concept underpins the result that our exercise aims to prove.In the context of the limit, the exercise asks us to show that as \( t \) approaches zero from the positive side, a weighted sum of powers \( \left( \sum_{i=1}^{n} c_{i} x_{i}^{t} \right)^{1/t} \) behaves like a product of powers. Using the properties of logarithms and limits helps to bridge the gap from a complex weighted sum to a more intuitive product of powers.
This transformation is essential because it turns a potential "infinite" expression into something concrete and understandable, thereby reconciling calculus concepts with algebraic expressions.