/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 56 Find the limits $$\lim _{x \righ... [FREE SOLUTION] | 91Ó°ÊÓ

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Find the limits $$\lim _{x \rightarrow \infty} x^{1 / \ln x}$$

Short Answer

Expert verified
The limit is \( e \).

Step by step solution

01

Rewrite the Expression

The given expression is \( x^{1/\ln x} \). To evaluate the limit as \( x \to \infty \), first, let's rewrite it using the exponential function: \( \text{exp}\left( \frac{1}{\ln x} \right)\ln(x) \).
02

Introduce a Substitution

Let \( y = \ln x \), which implies that as \( x \to \infty \), \( y \to \infty \) as well. The expression changes to \( \exp\left( \frac{y}{y} \right) \), where the inner expression simplifies to 1.
03

Evaluate the Exponential

Substitute back \( \exp(1) \) to reconfirm that the exponential of 1 is \( e \). Thus, the transformed expression becomes \( e \).
04

Conclude the Limit

Since the expression simplifies to \( e \) for \( x \to \infty \), the original limit is evaluated to \( e \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Limit at Infinity
When we talk about the "limit at infinity," we're interested in how a function behaves as its input value grows larger and larger without bound. In other words, we want to see what happens as the variable approaches infinity. It's a way to understand the long-term behavior of functions. This concept is crucial in calculus because it helps us see the end behavior of curves or graphs.
In the exercise given, we deal with the expression \( x^{1 / \ln x} \) as \( x \to \infty \). Our goal is to find out what value this expression approaches when \( x \) becomes extremely large. We achieve this by using calculus techniques to simplify the expression until it reaches a form that is easy to analyze and evaluate. Understanding limits at infinity is essential because they often describe the ultimate fate or value of processes modeled by functions, such as exponential growth and decay.
Exponential Functions
Exponential functions are those in which the variable appears in the exponent, for example, \( b^x \), where \( b \) is the base. These functions grow or decay at rates proportional to their current value, leading to rapid changes. Understanding exponential functions is fundamental in calculus, as they appear frequently in both natural phenomena and mathematical modeling.
In our exercise, we use the natural exponential function, \( \exp \), which is equivalent to raising \( e \) (approximately 2.718) to the power of a given expression. The step-by-step solution rewrites the initial expression \( x^{1/\ln x} \) as \( \exp\left( \frac{1}{\ln x} \right) \ln(x) \). This transformation is key because it switches the power-based expression into a more manageable exponential form.
By simplifying to an exponential form, we can then more easily evaluate the behavior of the function as \( x \) approaches infinity, culminating in the determination of the limit.
Logarithmic Substitution
Logarithmic substitution is a clever technique used to make expressions easier to handle, especially when dealing with limits and exponents. It involves substituting a variable with its logarithm, commonly transforming complex expressions into simpler forms.
In the problem, the substitution \( y = \ln x \) simplifies the expression significantly. As \( x \to \infty \), \( y \to \infty \) as well, making it easier to analyze the limit behavior since logarithms grow very slowly compared to polynomial or exponential functions.
  • The initial substitution turns the expression into \( \exp\left( \frac{y}{y} \right) \).
  • This simplifies to \( \exp(1) \), a very straightforward form.
This technique not only simplifies the evaluation but also unravels hidden structures in algebraic forms. Logarithmic substitution often uncovers patterns that aren't immediately obvious, assisting in limit evaluation.
Calculus Techniques
In solving calculus problems, especially limits, various techniques are at our disposal. These are essential for unraveling complexities in expressions that initially seem daunting.
  • One such technique, shown in the solution, is rewriting expressions to expose simpler forms or structures, such as converting \( x^{1/\ln x} \) into an exponential function.
  • Another is using substitutions, such as logarithmic substitution, to simplify the algebra and make it easier to analyze limits and other expressions.
By breaking down the original problem into manageable parts, we apply these calculations systematically to determine the result with clarity and accuracy.
These techniques are powerful because they transform complex calculus problems into easier steps, allowing us to discover the limiting behaviors, rates of change, and other dynamic properties of functions in both everyday applications and advanced mathematical contexts.

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a. When we cough, the trachea (windpipe) contracts to increase the velocity of the air going out. This raises the questions of how much it should contract to maximize the velocity and whether it really contracts that much when we cough. Under reasonable assumptions about the elasticity of the tracheal wall and about how the air near the wall is slowed by friction, the average flow velocity \(v\) can be modeled by the equation $$v=c\left(r_{0}-r\right) r^{2} \mathrm{cm} / \mathrm{sec}, \quad \frac{r_{0}}{2} \leq r \leq r_{0}$$ where \(r_{0}\) is the rest radius of the trachea in centimeters and \(c\) is a positive constant whose value depends in part on the length of the trachea. Show that \(v\) is greatest when \(r=(2 / 3) r_{0} ;\) that is, when the trachea is about \(33 \%\) contracted. The remarkable fact is that X-ray photographs confirm that the trachea contracts about this much during a cough. b. Take \(r_{0}\) to be 0.5 and \(c\) to be \(1,\) and graph \(v\) over the interval \(0 \leq r \leq 0.5 .\) Compare what you see with the claim that \(v\) is at a maximum when \(r=(2 / 3) r_{0}\).

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