Problem 49 - Calculus

Question

Use algebra to evaluate the limits. \(\lim _{h \rightarrow 0} \frac{\sqrt{4+h}-2}{h}[\text { Hint: Multiply by } \sqrt{4+h}+2\) in numerator and denominator. \(]\).

Step-by-Step Solution

Answer

The limit is \(\frac{1}{4}\).

Step 1: Identify the Problem

We need to evaluate the following limit by simplifying the expression:\[\lim _{h \rightarrow 0} \frac{\sqrt{4+h}-2}{h}\]A useful hint is given: multiply the numerator and the denominator by \( \sqrt{4+h}+2 \).

Step 2: Multiply by Conjugate

The expression is multiplied by the conjugate \( \sqrt{4+h}+2 \) in both the numerator and the denominator, transforming it to:\[\lim _{h \rightarrow 0} \frac{(\sqrt{4+h}-2)(\sqrt{4+h}+2)}{h(\sqrt{4+h}+2)}\]

Step 3: Apply the Difference of Squares

Using the difference of squares formula \((a-b)(a+b) = a^2 - b^2\), where \(a = \sqrt{4+h}\) and \(b = 2\), we simplify the numerator:\[\lim _{h \rightarrow 0} \frac{(4+h) - 4}{h(\sqrt{4+h}+2)} = \frac{h}{h(\sqrt{4+h}+2)}\]

Step 4: Simplify and Cancel Terms

Cancel out the \(h\) terms in the numerator and denominator:\[\lim _{h \rightarrow 0} \frac{1}{\sqrt{4+h}+2}\]

Step 5: Evaluate the Limit

Now, substitute \(h = 0\) to evaluate the limit:\[\lim _{h \rightarrow 0} \frac{1}{\sqrt{4+0}+2} = \frac{1}{2+2} = \frac{1}{4}\]The limit evaluates to \(\frac{1}{4}\).

Key Concepts

Difference of Squares

The difference of squares is a fundamental algebraic concept used to simplify certain expressions. It applies to expressions of the form \((a-b)(a+b)\), which simplifies to \(a^2-b^2\). This identity is incredibly useful when dealing with radical expressions like limits, especially before substituting values that could lead to undefined forms, such as division by zero.

In our limit evaluation problem, we start with the expression \(\sqrt{4+h}-2\). To simplify, we consider its conjugate \(\sqrt{4+h}+2\). By utilizing the difference of squares, we transform \((\sqrt{4+h}-2)(\sqrt{4+h}+2)\) into \((4+h) - 4\). This transformation neatly cancels out the radicand and leaves us with an expression easy to manage: \(h\).

Understanding and applying the difference of squares is essential, as it allows us to eliminate the square root and handle limits more gracefully.

Conjugate Method

The conjugate method is a powerful technique used in the evaluation of limits and rationalizing expressions. Specifically, when dealing with square roots, multiplying the numerator and the denominator by the conjugate can simplify the expression drastically.

Conjugates are expressions that differ only in the sign between two terms. For example, the conjugate of \(\sqrt{4+h} - 2\) is \(\sqrt{4+h} + 2\). Multiplying by the conjugate helps eliminate the square root from the numerator. This action is significant because it transforms a potentially indeterminate form by applying the difference of squares.

• The conjugate method ensures numerators become polynomials, facilitating easier manipulation.
• It prevents discontinuities or undefined forms due to division by zero, especially important while evaluating limits.

In our problem, by multiplying by the conjugate, we eventually reduce the expression into a more straightforward form \(\frac{h}{h(\sqrt{4+h}+2)}\), allowing cancellation and simplification.

Algebraic Simplification

Algebraic simplification is the process of rewriting mathematical expressions in a simpler form. This process is not only based on applying algebraic rules but also ensures expressions are manageable and easily evaluated.

After multiplying by the conjugate in a limit problem, simplification often follows by canceling common factors. In this particular exercise, after applying the difference of squares, we have the simplified expression \(\frac{h}{h(\sqrt{4+h}+2)}\).

• We notice that \(h\) appears in both the numerator and the denominator, which means they can be canceled out, provided \(h eq 0\).
• Once terms are canceled, the simplified form \(\frac{1}{\sqrt{4+h}+2}\) can be handled easily by substituting \(h = 0\).

This step is crucial in ensuring the limit can be evaluated, as it eliminates any problematic parts that could lead to undefined expressions, finally resulting in a clean, computable form.