Problem 67 - Thomas Calculus in SI Units

Question

L'Hópital's Rule does not help with the limits. 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: Identify indeterminate form

First, let's identify why L'Hôpital's Rule doesn't work. As \( x \to \infty \), both the numerator \( \sqrt{9x+1} \) and the denominator \( \sqrt{x+1} \) approach infinity. This implies an indeterminate form \( \frac{\infty}{\infty} \). However, applying L'Hôpital's Rule would cycle indefinitely because derivatives of both the numerator and denominator keep resulting in the same form.

Step 2: Simplify using algebraic manipulation

To simplify the expression, factor out \( \sqrt{x} \) from both the numerator and the denominator. This gives us: \[ \lim_{x \to \infty} \frac{\sqrt{9x+1}}{\sqrt{x+1}} = \lim_{x \to \infty} \frac{\sqrt{x} \cdot \sqrt{9 + \frac{1}{x}}}{\sqrt{x} \cdot \sqrt{1 + \frac{1}{x}}} \] This simplifies to: \[ \lim_{x \to \infty} \frac{\sqrt{9 + \frac{1}{x}}}{\sqrt{1 + \frac{1}{x}}} \]

Step 3: Evaluate the simplified limit

Since \( x \to \infty \), the terms \( \frac{1}{x} \) in both \( \sqrt{9 + \frac{1}{x}} \) and \( \sqrt{1 + \frac{1}{x}} \) become insignificant. Thus, the expression simplifies further to: \[ \lim_{x \to \infty} \frac{\sqrt{9}}{\sqrt{1}} = \frac{3}{1} = 3 \]

Key Concepts

Indeterminate Forms

Indeterminate forms often arise in calculus when evaluating limits, particularly when expressions take on an unclear form such as \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \). These forms signal that the direct evaluation of the limit is not straightforward and further analysis is necessary.
When you encounter an indeterminate form, it indicates that applying simple substitution won't yield clear results. In our case, both the numerator \( \sqrt{9x + 1} \) and the denominator \( \sqrt{x + 1} \) approach infinity as \( x \to \infty \), leading to the form \( \frac{\infty}{\infty} \).
Recognizing this indeterminate form is crucial because it helps guide your strategy for solving the problem. While some forms can be tackled using L'Hôpital's Rule, others may require alternative approaches such as algebraic manipulation. The aim is to transform the expression into a form that allows clear limit evaluation.

Algebraic Manipulation

Algebraic manipulation involves reorganizing and simplifying expressions to make them easier to interpret and solve. In the case of limit problems, this often means factoring, expanding, or rationalizing to eliminate complexities.
For the given problem, we can simplify \( \lim_{x \to \infty} \frac{\sqrt{9x+1}}{\sqrt{x+1}} \) by factoring out \( \sqrt{x} \) from both the numerator and the denominator. This step reshapes the original function into:
\[ \lim_{x \to \infty} \frac{\sqrt{x} \cdot \sqrt{9 + \frac{1}{x}}}{\sqrt{x} \cdot \sqrt{1 + \frac{1}{x}}} \]
By canceling out common terms, you simplify the limit expression further to \( \lim_{x \to \infty} \frac{\sqrt{9 + \frac{1}{x}}}{\sqrt{1 + \frac{1}{x}}} \).
Demystifying the expression through algebraic manipulation not only reveals the underlying simplicity of the problem but also makes it easier to identify any limiting behavior, paving the way for smooth evaluation.

L'Hôpital's Rule

L'Hôpital's Rule is a handy tool for resolving limits that result in indeterminate forms like \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \). The rule states that if these forms are present, the limit of a quotient of functions can be found by differentiating the numerator and the denominator separately, and then taking the limit of the resulting expression.
However, L'Hôpital's Rule isn't always the most efficient route. In the given exercise, applying the rule would repeatedly result in the same indeterminate form \( \frac{\infty}{\infty} \). Thus, it doesn't resolve the problem when simple algebraic manipulation does it more effectively.
It's important to recognize when to use L'Hôpital's Rule and when to resort to other strategies. If derivatives don't simplify the equation or cycle endlessly, exploring methods like factorization or rationalization might be more insightful.
Understanding when and how to apply this rule saves time and leads to solutions more efficiently when tackling calculus problems.