Problem 25 - Calculus: Early Transcendentals

Question

Compute the following limits. $$ \lim _{x \rightarrow 0} \frac{\sqrt{x+1}-1}{\sqrt{x+4}-2} $$

Step-by-Step Solution

Answer

The limit is \( \frac{1}{2} \).

Step 1: Determine Indeterminate Form

First, substitute the value of \( x \rightarrow 0 \) into the expression. We get \( \frac{\sqrt{0+1}-1}{\sqrt{0+4}-2} \). This simplifies to \( \frac{1-1}{2-2} = \frac{0}{0} \), indicating an indeterminate form.

Step 2: Rationalize the Numerator and Denominator

To eliminate the indeterminate form, multiply the numerator and the denominator by the conjugate of each expression: \( \lim_{x \to 0} \frac{(\sqrt{x+1} - 1)(\sqrt{x+1} + 1)}{(\sqrt{x+4} - 2)(\sqrt{x+4} + 2)} \). This simplifies to \( \frac{x}{x} \cdot \frac{\sqrt{x+1} + 1}{\sqrt{x+4} + 2} \).

Step 3: Simplify the Expression

After applying the conjugate, the expression simplifies to \( \lim_{x \to 0} \frac{x}{x} \cdot \frac{\sqrt{x+1} + 1}{\sqrt{x+4} + 2} \). Cancel the \( x \) terms to simplify it to \( \lim_{x \to 0} \frac{\sqrt{x+1} + 1}{\sqrt{x+4} + 2} \).

Step 4: Substitute the Limit

Now substitute \( x = 0 \) directly into the simplified expression: \( \frac{\sqrt{0+1} + 1}{\sqrt{0+4} + 2} = \frac{1 + 1}{2 + 2} = \frac{2}{4} \).

Step 5: Simplify the Result

Simplify the fraction obtained: \( \frac{2}{4} = \frac{1}{2} \).

Step 6: Conclusion

Thus, the limit of the given expression as \( x \) approaches 0 is \( \frac{1}{2} \).

Key Concepts

Rationalization technique

When faced with a limit problem that results in an indeterminate form like \( \frac{0}{0} \), a useful method to resolve this is the rationalization technique. This technique involves multiplying the expression by a "conjugate." The goal is to eliminate the square roots from numerator and denominator.

For instance, consider an expression \( \frac{\sqrt{x+1}-1}{\sqrt{x+4}-2} \). You can multiply both the numerator and the denominator by the conjugate \( \sqrt{x+1}+1 \) and \( \sqrt{x+4}+2 \) to rationalize both parts. After performing the operation, the new expression becomes simpler.

- The conjugate helps clear radical expressions by utilizing the difference of squares formula: \( (a-b)(a+b) = a^2 - b^2 \). - This approach simplifies the limit and often allows direct substitution to calculate the final result.

Indeterminate forms

In calculus, limits often initially appear in an indeterminate form, such as \( \frac{0}{0} \). This means direct substitution gives an undefined result. These forms require further manipulation to find the actual limit.

- Indeterminate forms typically arise in limits involving fractions, radicals, or logarithmic expressions.
- Through techniques like factoring, rationalization, or L'Hôpital's Rule, we can rewrite the expression to find a well-defined limit.

In our exercise, direct substitution of \( x = 0 \) results in \( \frac{0}{0} \). By applying the rationalization technique, we can simplify the expression, transcending beyond the indeterminacy to reach a concrete result.

Conjugates in calculus

A conjugate in calculus is used primarily to simplify expressions especially when dealing with radicals or irrational numbers. The conjugate of an expression like \( a-b \) is \( a+b \). When multiplied together, they eliminate the radical through the difference of squares formula, \( a^2 - b^2 \).

For example, take \( \sqrt{x+1} - 1 \). Its conjugate is \( \sqrt{x+1} + 1 \). Multiplying these results in \( (x+1) - 1^2 \), thereby removing the square root.

- Conjugates stabilize indeterminate forms and simplify the algebraic process.
- This is crucial in calculus as it allows us to more simply evaluate limits, derivatives, and integrals involving radicals.

Simplifying expressions

Simplifying expressions is often necessary to evaluate limits effectively. After using techniques like rationalization, it's important to continue simplifying until the expression becomes manageable for direct substitution.

In our exercise, once the rationalization removes roots, we simplify \( \frac{x}{x} \) to cancel out terms and eventually substitute \( x \) with its limit value. This leads to a simplified expression free from initial indeterminate forms.

- Simplification involves canceling out common terms or further breaking down complex expressions.
- It results in reduced expressions that readily yield their limits.

With a simplified form, our exercise ultimately shows \( \frac{\sqrt{x+1} + 1}{\sqrt{x+4} + 2} \). Substituting \( x = 0 \) gives \( \frac{2}{4} \), illustrating how simplification eases calculation and unveils the result efficiently.