Problem 28 - Calculus: Early Transcendentals

Question

Compute the following limits. $$ \lim _{x \rightarrow 0} \frac{\sqrt{x^{2}+1}-1}{\sqrt{x+1}-1} $$

Step-by-Step Solution

Answer

The limit is 0.

Step 1: Understand the Problem

We need to find the limit as \( x \) approaches 0 for the expression \( \frac{\sqrt{x^{2}+1}-1}{\sqrt{x+1}-1} \). Since both the numerator and denominator approach 0 as \( x \to 0 \), this results in an indeterminate form \( \frac{0}{0} \). We use algebraic manipulation to simplify this expression.

Step 2: Rationalize the Numerator

To eliminate the square root from the numerator, multiply the numerator and the denominator by the conjugate of the numerator, \( \sqrt{x^2+1}+1 \). This gives:\[\frac{(\sqrt{x^{2}+1}-1)(\sqrt{x^{2}+1}+1)}{(\sqrt{x+1}-1)(\sqrt{x^{2}+1}+1)} = \frac{x^{2}}{(\sqrt{x+1}-1)(\sqrt{x^{2}+1}+1)}\] since \((a-b)(a+b) = a^2 - b^2\).

Step 3: Rationalize the Denominator

Now, rationalize the denominator by multiplying both the numerator and the denominator by the conjugate of the denominator, \( \sqrt{x+1}+1 \). This results in:\[\frac{x^{2}(\sqrt{x+1}+1)}{((\sqrt{x+1})^2-1)(\sqrt{x^{2}+1}+1)} = \frac{x^{2}(\sqrt{x+1}+1)}{x(\sqrt{x^{2}+1}+1)}\] because \((a-b)(a+b) = a^2 - b^2\) simplifies to \(x\) for the denominator.

Step 4: Simplify the Expression

Cancel the common \( x \) factor from the numerator and denominator:\[\frac{x(\sqrt{x+1}+1)}{\sqrt{x^{2}+1}+1}\]Now substitute \( x = 0 \) into the simplified expression.

Step 5: Evaluate the Limit

Substitute \( x = 0 \) in the simplified expression:\[\lim _{x \to 0} \frac{x(\sqrt{x+1}+1)}{\sqrt{x^{2}+1}+1} = \frac{0(\sqrt{0+1}+1)}{\sqrt{0^{2}+1}+1} = \frac{0 \cdot 2}{2} = 0\]Thus, the limit of the expression as \( x \to 0 \) is 0.

Key Concepts

Indeterminate Forms

In calculus, indeterminate forms take shape when evaluating limits that do not appear directly solvable, such as \( \frac{0}{0} \). These forms signal that we cannot determine the limit's value solely by substitution. Instead, it indicates that further algebraic or analytic manipulation is necessary to resolve the limit. When faced with indeterminate forms, strategies like factoring, rationalizing, or using L'Hôpital's Rule become valuable tools. These techniques transform the expression into a form where the limits can be directly calculated, steering away from the uncertainty of the indeterminate form.

Algebraic Manipulation

Algebraic manipulation involves reworking an expression to simplify or alter it in order to find a limit or solve an equation. In the given problem, manipulation is crucial as the original expression results in an indeterminate form.

• Step 1: Factor or multiply by a conjugate to simplify fractions or radicals.
• Step 2: Cancel common terms, ensuring no restrictions are ignored.

Using algebraic manipulation helps in transforming complex or awkward expressions into ones more conducive to limit evaluation or further mathematical analysis. It streamlines the problem into solvable parts.

Conjugate Multiplication

Conjugate multiplication is a technique used to eliminate square roots from the numerator or denominator of a fraction. When dealing with radicals, multiplying by the conjugate employs the identity \((a-b)(a+b) = a^2 - b^2\).

• Original: \( \sqrt{x^2+1}-1 \)
• Conjugate: \( \sqrt{x^2+1}+1 \)

This step is pivotal in rationalizing expressions, simplifying them without changing their value. It turns potentially problematic radicals into simpler polynomials, facilitating easier substitution in limit calculations.

Limit Evaluation

Limit evaluation comes after simplifying the original expression through techniques like conjugate multiplication and canceling terms. The goal is to substitute the variable's approaching value to find the limit. Once the expression is simplified to avoid the indeterminate form, we substitute the limit point.

• Ensure all terms are reduced appropriately so straightforward substitution is possible.
• Re-evaluate terms that may still pose issues post-simplification.

In this problem, after simplification, substituting \(x = 0\) showed no undefined expressions, leading to an easy evaluation: the limit is \(0\). This step confirms the calculated value matches the behavior of the function as it approaches the specified point.