Problem 45 - Calculus for Biology and Medicine

Question

In Problems \(29-48\), find the limits. $$ \lim _{x \rightarrow 0} \frac{\sqrt{x^{2}+9}-3}{x^{2}} $$

Step-by-Step Solution

Answer

The limit is \( \frac{1}{6} \).

Step 1: Identify the Limit Type

Observe the limit expression \( \lim _{x \rightarrow 0} \frac{\sqrt{x^{2}+9}-3}{x^{2}} \). Notice that as \( x \rightarrow 0 \), both the numerator and the denominator approach 0, leading to a \( \frac{0}{0} \) indeterminate form. Thus, algebraic manipulation or L'Hôpital's Rule can be applied.

Step 2: Rationalize the Numerator

Multiply the numerator and denominator by the conjugate of the numerator, \( \sqrt{x^{2}+9}+3 \), to rationalize it:\[\lim _{x \rightarrow 0} \frac{(\sqrt{x^{2}+9}-3)(\sqrt{x^{2}+9}+3)}{x^{2}(\sqrt{x^{2}+9}+3)}\]The numerator becomes a difference of squares: \( ((\sqrt{x^{2}+9})^2 - 3^2) = x^{2} + 9 - 9 = x^{2} \).

Step 3: Simplify the Expression

Simplify the expression by canceling \( x^{2} \) from the numerator and the denominator:\[\lim _{x \rightarrow 0} \frac{x^{2}}{x^{2}(\sqrt{x^{2}+9}+3)} = \lim _{x \rightarrow 0} \frac{1}{\sqrt{x^{2}+9}+3}\]

Step 4: Evaluate the Limit

Substitute \( x = 0 \) into the simplified expression. The resulting expression is:\[\lim _{x \rightarrow 0} \frac{1}{\sqrt{0^{2}+9}+3} = \frac{1}{3+3} = \frac{1}{6}\]

Step 5: Conclude

Since the expression has been evaluated to a finite value without any indeterminate form, the limit is:\[\lim _{x \rightarrow 0} \frac{\sqrt{x^{2}+9}-3}{x^{2}} = \frac{1}{6}\]

Key Concepts

L'Hôpital's Rule

When encountering a limit problem like \( \lim _{x \rightarrow 0} \frac{\sqrt{x^{2}+9}-3}{x^{2}} \) that results in an indeterminate form \( \frac{0}{0} \), L'Hôpital's Rule can be a useful tool to solve it. L'Hôpital's Rule is applied to indeterminate forms \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \) and is stated as follows:

• Differentiate the numerator and the denominator separately.
• Re-evaluate the limit of the resulting function.

In this exercise, although the rule could be applied, it is more effectively solved through an algebraic technique called rationalizing. However, it demonstrates when L'Hôpital's Rule may be considered as a possible method for evaluating limits when you encounter these specific types of indeterminate forms.

Rationalizing Technique

To solve the given limit \( \lim _{x \rightarrow 0} \frac{\sqrt{x^{2}+9}-3}{x^{2}} \), we use the rationalizing technique, which is an algebraic manipulation involving the conjugate. The goal is to eliminate the square root in the numerator by multiplying both the numerator and denominator by the conjugate of the numerator, which is \( \sqrt{x^2 + 9} + 3 \). When you multiply:

• The numerator becomes a difference of squares: \((\sqrt{x^2 + 9})^2 - 3^2 \), simplifying to \( x^2 \).
• The denominator becomes \( x^2(\sqrt{x^2 + 9} + 3) \), allowing further simplification.

After multiplying and simplifying the expression by canceling \( x^2 \) from both the numerator and denominator, we get \( \lim _{x \rightarrow 0} \frac{1}{\sqrt{x^2 + 9} + 3} \), which is readily evaluated by direct substitution. The rationalizing technique is particularly useful for dealing with limits involving radicals.

Indeterminate Forms

Indeterminate forms, such as \( \frac{0}{0} \), \( \frac{\infty}{\infty} \), \( 0 \times \infty \), can make evaluating limits challenging. These forms occur when the limit produces no straightforward numerical value. Let's explore the \( \frac{0}{0} \) indeterminate form:

• In the problem \( \lim _{x \rightarrow 0} \frac{\sqrt{x^2 + 9} - 3}{x^2} \), substituting \( x = 0 \) directly into the expression results in \( \frac{0}{0} \).
• This signals that the limit cannot be determined merely by substitution and requires further analysis.

Approaches to resolve indeterminate forms include L'Hôpital's Rule or algebraic techniques, such as rationalizing. Recognizing these forms is crucial because it informs you that the problem needs a deeper investigation to evaluate the limit correctly.