Problem 67 - Thomas Calculus

Question

L'Hopital's Rule does not help with the limits in Exercises \(67-74 .\) Try it- you just keep on cycling. Find the limits some other way. $$ \lim _{x \rightarrow \infty} \frac{\sqrt{9 x+1}}{\sqrt{x+1}} $$

Step-by-Step Solution

Answer

The limit is 3.

Step 1: Recognize indeterminate form

As \(x\) approaches infinity, both the numerator \(\sqrt{9x+1}\) and the denominator \(\sqrt{x+1}\) trend toward infinity, leading to an indeterminate form \(\frac{\infty}{\infty}\). At first glance, this might suggest trying L'Hopital's Rule, but instead, let's simplify the expression directly.

Step 2: Simplify the expression

Factor out \(\sqrt{x}\) from both the numerator and the denominator:\[\frac{\sqrt{9x+1}}{\sqrt{x+1}} = \frac{\sqrt{x(9 + \frac{1}{x})}}{\sqrt{x(1 + \frac{1}{x})}}\]This reduces to:\[\frac{\sqrt{x}\sqrt{9 + \frac{1}{x}}}{\sqrt{x}\sqrt{1 + \frac{1}{x}}} = \frac{\sqrt{9 + \frac{1}{x}}}{\sqrt{1 + \frac{1}{x}}}\]

Step 3: Evaluate the limit

As \(x\) approaches infinity, both \(\frac{1}{x}\) terms in the numerator and denominator approach 0. Thus, the expression simplifies to:\[\frac{\sqrt{9 + 0}}{\sqrt{1 + 0}} = \frac{\sqrt{9}}{\sqrt{1}} = \frac{3}{1} = 3\]Therefore, the limit is 3.

Key Concepts

Indeterminate Form

When dealing with limits, you might encounter expressions that are in an indeterminate form. This is a situation where the limit expression doesn't lead you straight to a clear and definite answer. An example of this is \( \frac{\infty}{\infty} \). In our exercise, as \( x \) approaches infinity, both the numerator \( \sqrt{9x+1} \) and the denominator \( \sqrt{x+1} \) become very large, specifically, they both tend towards infinity. This results in the indeterminate form \( \frac{\infty}{\infty} \).

When you come across such a form, tools like L'Hopital's Rule might be tempting to use. However, L'Hopital's Rule requires that the derivatives in consecutive rounds don't just keep producing more of the same indeterminate forms. Here, direct simplification often yields a clearer path to the solution than repeated differentiation.

Limit Evaluation

Limit evaluation is a central concept in calculus that involves finding the value that a function approaches as the input (usually \( x \)) approaches a particular point. In our example, the key is to evaluate the limit as \( x \to \infty \). The first step is to simplify the expression if possible.

By factoring out \( \sqrt{x} \) from both the numerator and the denominator, you can see the power of simplification:

• Numerator: \( \sqrt{9x+1} = \sqrt{x(9 + \frac{1}{x})} \).
• Denominator: \( \sqrt{x+1} = \sqrt{x(1 + \frac{1}{x})} \).

This simplification removes the \( x \) term:
\[ \frac{\sqrt{9 + \frac{1}{x}}}{\sqrt{1 + \frac{1}{x}}} \]
As \( x \) approaches infinity, \( \frac{1}{x} \) approaches 0. Thus, the expression simplifies further:\[ \frac{\sqrt{9}}{\sqrt{1}} = \frac{3}{1} = 3 \]
So, the limit evaluates to 3.

Infinity in Calculus

Infinity in calculus is a concept that frequently represents the behavior of functions as they extend indefinitely. When approaching infinity in calculations, \( x \to \infty \) or negative infinity \( x \to -\infty \), we often deal with limits that describe the end behavior of functions.

Our exercise involved finding the limit of a rational expression as \( x \) approached infinity. Calculus provides us with the tools to handle these scenarios elegantly, understanding the growth rates of functions.

For the given problem, both the numerator and the denominator independently tend towards infinity. Simplifying by looking at the rates of growth via factor and cancel gives us a path to solve this indeterminate situation. Such insights are crucial in capturing what happens in an expression as we push boundaries towards infinity, and it's a fundamental aspect in calculus. Erasing terms that become negligible, like \( \frac{1}{x} \) going towards 0, lights the way to the correct answer as we derived a finite result from an initial infinite dynamic.