Problem 16 - Thomas Calculus

Question

Use 1'Hopital's Rule to find the limits in Exercises \(7-26\). $$ \lim _{x \rightarrow 2} \frac{\sqrt{x^{2}+5}-3}{x^{2}-4} $$

Step-by-Step Solution

Answer

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

Step 1: Verify Indeterminate Form

First, substitute the value of \( x = 2 \) into the given expression \( \frac{\sqrt{x^{2}+5}-3}{x^{2}-4} \). Numerator: \( \sqrt{2^2 + 5} - 3 = \sqrt{9} - 3 = 3 - 3 = 0 \). Denominator: \( 2^2 - 4 = 4 - 4 = 0 \). Since both numerator and denominator evaluate to zero, the expression is in an indeterminate form \( \frac{0}{0} \).

Step 2: Apply L'Hopital's Rule

Since the limit is of the indeterminate form \( \frac{0}{0} \), we can apply L'Hopital's Rule. Take the derivative of the numerator and the derivative of the denominator separately.

Step 3: Differentiate the Numerator

The numerator is \( \sqrt{x^2+5} - 3 \). Differentiating \( \sqrt{x^2+5} \) we use the chain rule: \( \frac{d}{dx}[\sqrt{x^2+5}] = \frac{1}{2\sqrt{x^2+5}} \cdot 2x = \frac{x}{\sqrt{x^2+5}} \). The derivative of \(-3\) is \(0\). Therefore, the derivative of the numerator is \( \frac{x}{\sqrt{x^2+5}} \).

Step 4: Differentiate the Denominator

The denominator is \( x^2 - 4 \). The derivative is \( 2x \).

Step 5: Evaluate the Limit

Using the derivatives found, the expression becomes \( \lim_{x \rightarrow 2} \frac{\frac{x}{\sqrt{x^2+5}}}{2x} = \lim_{x \rightarrow 2} \frac{x}{2x\sqrt{x^2+5}} = \lim_{x \rightarrow 2} \frac{1}{2\sqrt{x^2+5}} \). Substituting \( x = 2 \), we get \( \frac{1}{2\sqrt{9}} = \frac{1}{6} \).

Key Concepts

Indeterminate Form

When dealing with limits in calculus, you might encounter something called an "indeterminate form." An indeterminate form arises when you substitute a value into a limit expression, and it results in an ambiguous form like \( \frac{0}{0} \). This doesn't mean the limit doesn't exist; instead, it signals that more work is needed to find the actual limit.
For our specific problem, when substituting \( x = 2 \) into the expression \( \frac{\sqrt{x^{2}+5}-3}{x^{2}-4} \), both the numerator and denominator result in 0. This creates the indeterminate form \( \frac{0}{0} \). This is a classic scenario where using L'Hôpital's Rule is appropriate, as it helps us properly evaluate the limit.

Derivatives

Derivatives are a fundamental concept in calculus, helping us understand the rate of change of functions. They are essential when using L'Hôpital's Rule, which involves taking derivatives to resolve indeterminate forms.
In the given exercise, after identifying the indeterminate form, the next step is to compute the derivatives of both the numerator and the denominator.

• The numerator's derivative: Start with \( \sqrt{x^2+5}-3 \). Here, the chain rule applies, leading to: \( \frac{x}{\sqrt{x^2+5}} \).
• The derivative of \(-3\) is simply 0, since it's a constant.

• The denominator's derivative: Differentiate \( x^2-4 \) to obtain \( 2x \).

Using these derivatives, we reformulate the original limit, enabling us to evaluate it without the indeterminate form issue.

Limits

Limits in calculus describe the behavior of functions as they approach a certain point or infinity. They are crucial for understanding how functions behave at specific values, particularly when direct substitution isn't possible due to undefined forms.
To determine the limit for our problem, we apply L'Hôpital's Rule after computing the derivatives. This transforms the original expression, allowing us to compute:

• \( \lim_{x \rightarrow 2} \frac{\frac{x}{\sqrt{x^2+5}}}{2x} \), which further simplifies to \( \lim_{x \rightarrow 2} \frac{1}{2\sqrt{x^2+5}} \).

After simplifying, we substitute \( x = 2 \) directly into the simplified expression, yielding \( \frac{1}{6} \).
This process highlights how limits, when combined with derivatives through L'Hôpital's Rule, enable us to evaluate expressions that are initially undefined at a specific point.