Problem 34 - Essential Calculus

Question

In Exercises \(15-36,\) find the limit. $$ \lim _{x \rightarrow \infty}\left[\frac{5}{2}+\ln \left(\frac{x^{2}+1}{x^{2}}\right)\right] $$

Step-by-Step Solution

Answer

\(\frac{5}{2}\)

Step 1: Identify Individual Terms

Firstly, identify the individual terms in the function that the limit is being taken of. These are \(\frac{5}{2}\) and \(\ln\left(\frac{x^{2}+1}{x^{2}}\right)\). The limit of the first term doesn't depend on x, so it remains \(\frac{5}{2}\). The second term is a natural logarithm of a ratio of two functions of x, and this will be treated separately.

Step 2: Evaluate the Limit of the Second Term

Let's consider the function inside the logarithm, \(\frac{x^{2}+1}{x^{2}}\). As x approaches infinity, the \(+1\) in the numerator becomes negligibly small compared with \(x^{2}\), so the limit of this term can be simplified to \(\frac{x^{2}}{x^{2}} = 1\). The natural logarithm of 1 is 0, so the limit of the entire second term as x approaches infinity is 0.

Step 3: Combine the Limits of the Terms

Now that the limit of the second term was found to be 0, combine this with the limit of the first term, \(\frac{5}{2}\). Because the limit of a sum of functions is the sum of their limits, the limit of the entire function as x approaches infinity is therefore \(\frac{5}{2} + 0 = \frac{5}{2}\).

Key Concepts

Understanding the Natural Logarithm

The natural logarithm, denoted as \( \ln(x) \), is a function that's widely used in various branches of mathematics, including calculus, to unravel the mysteries of exponential growth and decay. Its special property lies in the base of the logarithm, which is the irrational number 'e' (approximately 2.71828). This number is particularly significant because it creates a unique relationship where the rate of growth (or decay) is proportional to the value of the function at any point.

In essence, when you see \( \ln(x) \), imagine it as asking the question: 'To what power must we raise 'e' to achieve the value x?' It’s essential to understand that the natural logarithm of 1 is always 0, and the function is undefined for values less than or equal to zero. This is because no power of 'e' will result in a negative number or zero.

Demystifying Limits at Infinity

Diving into limits at infinity, we engage with the behavior of functions as the variable approaches an infinitely large value, or \( x \rightarrow \infty \). It's like watching a spaceship launch — as it moves farther from Earth (\( x \)), we want to know if it travels towards a specific star (limit) or just continues into the vastness of space without direction.

Specifically, we look at the end behavior of the function — does it level off to a horizontal line, or does it keep climbing or falling without bound? In the case of the ratio \( \frac{x^{2}+1}{x^{2}} \), as \( x \rightarrow \infty \), the terms involving the highest power of x in the numerator and denominator dominate the behavior of the ratio, reducing it to \( \frac{x^{2}}{x^{2}} = 1 \). Thus, despite \( x \) growing infinitely large, the limit of this ratio simplifies to a certainty, not an endless journey.

Applying Limit Properties

Limit properties are like the rules of the road for navigating functions on their path towards a limit. They are the shortcuts that allow us to simplify complex problems by breaking them down into more manageable pieces. Specifically, they give us permission to take limits term by term when dealing with sums, differences, products, and quotients of functions, as appropriate.

For example, in the expression \( \lim _{x \rightarrow \infty}\left[\frac{5}{2} + \ln \left(\frac{x^{2}+1}{x^{2}}\right)\right] \), we utilize the property that the limit of the sum is the sum of the limits. Knowing that the limit of a constant term remains unchanged, and the limit of the \( \ln \) function of our particular ratio goes to 0, we neatly wrap up the problem into a single, finite value — quasi a package with a definitive label, rather than one with an ever-increasing tracking number.