Problem 62 - Thomas Calculus

Question

Find the limits in Exercises \(51-66\) $$ \lim _{x \rightarrow \infty}\left(\frac{x^{2}+1}{x+2}\right)^{1 / x} $$

Step-by-Step Solution

Answer

The limit is 1.

Step 1: Identify the Type of Limit

We have the expression \( \lim_{x \rightarrow \infty}\left(\frac{x^{2}+1}{x+2}\right)^{1/x} \). This is an indeterminate form \(1^{\infty}\), making this a good candidate for using logarithms.

Step 2: Apply Natural Logarithm

To handle the indeterminate form, take the natural logarithm: \( y = \ln\left(\left(\frac{x^{2}+1}{x+2}\right)^{1/x}\right) \) Simplify it to \( y = \frac{1}{x}\ln\left(\frac{x^{2}+1}{x+2}\right) \).

Step 3: Simplify the Logarithmic Expression

Further simplify: \( y = \frac{1}{x}\left(\ln(x^{2}+1) - \ln(x+2)\right) \), which can be rewritten as \( y = \frac{1}{x} \ln(x^{2}+1) - \frac{1}{x} \ln(x+2) \).

Step 4: Find Each Component's Limit

First evaluate \( \lim_{x \rightarrow \infty} \frac{1}{x} \ln(x^{2}+1) \). Approximating, \( \ln(x^{2}+1) \sim \ln(x^{2}) = 2\ln(x) \), so it approximates to \( \frac{2\ln(x)}{x} \). Since \( \ln(x)/x \rightarrow 0 \) as \( x \rightarrow \infty \), \( \frac{2\ln(x)}{x} \rightarrow 0 \).

Step 5: Evaluate the Second Key Limit

Next, evaluate \( \lim_{x \rightarrow \infty} \frac{1}{x} \ln(x+2) \). Again, approximating using \( \ln(x+2) \sim \ln(x) \), we have \( \frac{\ln(x)}{x} \rightarrow 0 \).

Step 6: Combine Results and Exponentiate

Since both components tend towards 0, the entire expression \( y \rightarrow 0 \). Thus, the original limit \[ \lim_{x \rightarrow \infty}\left(\frac{x^{2}+1}{x+2}\right)^{1/x} = e^y = e^0 = 1 \].

Key Concepts

Indeterminate Forms

When dealing with limits, you might encounter expressions that seem to trail in two or more directions, making the limit unclear at first glance. These are known as indeterminate forms. The most common types are

• \(\frac{0}{0}\)
• \(\frac{\infty}{\infty}\)
• \(\infty - \infty\)
• \(0 \cdot \infty\)
• \(1^\infty\)
• \(0^0\)
• \(\infty^0\)

In this exercise, the expression \(\left(\frac{x^2+1}{x+2}\right)^{1/x}\) is an indeterminate form of \(1^{\infty}\). Such forms are particularly tricky because they involve competing growth rates of the numerator and the denominator, or, in this case, the base and the exponent. Recognizing an indeterminate form is crucial for deciding the method of evaluation, often involving manipulating the expression with algebra or calculus techniques.

Logarithmic Simplification

Using logarithms to simplify complex expressions is a common strategy, especially when dealing with indeterminate forms. By applying the natural logarithm, we take advantage of its property to transform multiplicative processes into additive ones.For the problem at hand:

• We started with \(\ln\left(\left(\frac{x^2+1}{x+2}\right)^{1/x}\right)\)
• This becomes \(\frac{1}{x}\ln\left(\frac{x^2+1}{x+2}\right)\)
• Using the property \(\ln(\frac{a}{b}) = \ln(a) - \ln(b)\), it simplifies further to \(\frac{1}{x}(\ln(x^2+1) - \ln(x+2))\)

These steps are crucial for breaking down the expression into manageable parts to determine the limit.

Natural Logarithm

The natural logarithm, denoted as \(\ln\), is a mathematical function with unique properties. It is the logarithm to the base \(e\), where \(e\) is approximately 2.718. The natural logarithm is continuous, differentiable, and has a straightforward derivative: \(\frac{d}{dx} \ln(x) = \frac{1}{x}\).This makes it very useful in calculus because it elegantly handles multiplicative relationships and converts them into additive ones. As seen in the exercise, \(y = \frac{1}{x}(\ln(x^2+1) - \ln(x+2))\), the natural logarithm allowed us to transform a complex power into a simpler dilational form. This transformation is what sets the stage for further calculus techniques, such as evaluating limits.

Limits at Infinity

Limits at infinity focus on understanding the behavior of a function as the variable approaches infinity. The challenge here arises from needing to determine the ultimate trend or behavior of a function as it stretches towards unbounded growth.In our exercise, we simplify the separate terms of the logs:

• Approximating \(\ln(x^2+1)\) by \(2\ln(x)\) and \(\ln(x+2)\) by \(\ln(x)\).
• Each individual term \(\frac{1}{x}\ln(x)\) decays towards 0 as \(x\) grows large because the logarithm grows much slower than the linear term in the denominator.

The convergence of both terms to zero allows us to simplify the original limit to \(e^0\), resulting in the answer of 1. Analyzing limits at infinity helps identify end behaviors, which are crucial in understanding polynomial or rational expressions' dominance in given contexts.