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Ch. 7 - Logarithmic, Exponential Functions, and Hyperbolic Functions
Briggs - Calculus: Early Transcendentals 3rd Edition
Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243Not the one you use?Change textbook
Chapter 7, Problem 7.R.29b

Log-normal probability distribution A commonly used distribution in probability and statistics is the log-normal distribution. (If the logarithm of a variable has a normal distribution, then the variable itself has a log-normal distribution.) The distribution function is
f(x) = 1/xσ√(2π) e⁻ˡⁿ^² ˣ / ²σ^², for x ≥ 0
where ln x has zero mean and standard deviation σ > 0.
b. Evaluate lim x → 0 ƒ(x). (Hint: Let x = eʸ.)

Verified step by step guidance
1
Start by writing down the given probability density function (pdf) for the log-normal distribution: \(f(x) = \frac{1}{x \sigma \sqrt{2\pi}} e^{-\frac{(\ln x)^2}{2 \sigma^2}}\), for \(x \geq 0\).
To evaluate the limit as \(x\) approaches 0, use the hint and substitute \(x = e^y\). This means as \(x \to 0^+\), we have \(y = \ln x \to -\infty\).
Rewrite the function in terms of \(y\): \(f(e^y) = \frac{1}{e^y \sigma \sqrt{2\pi}} e^{-\frac{y^2}{2 \sigma^2}} = \frac{1}{\sigma \sqrt{2\pi}} e^{-y} e^{-\frac{y^2}{2 \sigma^2}}\).
Combine the exponents to get a single exponential expression: \(f(e^y) = \frac{1}{\sigma \sqrt{2\pi}} e^{-y - \frac{y^2}{2 \sigma^2}}\).
Analyze the behavior of the exponent \(-y - \frac{y^2}{2 \sigma^2}\) as \(y \to -\infty\). This will help determine the limit of \(f(x)\) as \(x \to 0^+\).

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Log-normal Distribution

A log-normal distribution describes a random variable whose logarithm is normally distributed. If Y = ln(X) follows a normal distribution, then X is log-normally distributed. This distribution is skewed and only defined for positive values, commonly used in modeling multiplicative processes.
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Limit Evaluation Using Substitution

Evaluating limits involving complex functions often requires substitution to simplify the expression. Here, substituting x = e^y transforms the limit as x approaches 0 into a limit as y approaches -∞, making it easier to analyze the behavior of the function near zero.
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Behavior of Exponential and Logarithmic Functions at Infinity

Understanding how exponential and logarithmic functions behave as their arguments approach infinity or negative infinity is crucial. For example, e^y approaches 0 as y → -∞, and ln(x) approaches -∞ as x → 0+, which helps in evaluating limits involving these functions.
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