Problem 22 - Calculus Early Transcendentals

Question

Evaluate the limit, if it exists. $$\lim _{n \rightarrow 0} \frac{\sqrt{1+h}-1}{h}$$

Step-by-Step Solution

Answer

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

Step 1: Identify the Form

First, observe that as \( h \rightarrow 0 \), both the numerator and the denominator approach zero. Therefore, this limit is an indeterminate form \( \frac{0}{0} \), suggesting that we need to manipulate the expression to evaluate the limit.

Step 2: Rationalize the Numerator

To resolve the indeterminate form, multiply the numerator and the denominator by the conjugate of the numerator, \( \sqrt{1+h}+1 \). This gives:\[\lim_{h \to 0} \frac{\sqrt{1+h}-1}{h} \cdot \frac{\sqrt{1+h}+1}{\sqrt{1+h}+1} = \lim_{h \to 0} \frac{(\sqrt{1+h}-1)(\sqrt{1+h}+1)}{h(\sqrt{1+h}+1)}\]

Step 3: Simplify the Expression

Using the difference of squares formula, convert the numerator:\[(\sqrt{1+h}-1)(\sqrt{1+h}+1) = (1+h) - 1^2 = h\]Substitute back into the expression:\[\lim_{h \to 0} \frac{h}{h(\sqrt{1+h}+1)} = \lim_{h \to 0} \frac{1}{\sqrt{1+h}+1}\]

Step 4: Evaluate the Limit

With the expression simplified, substitute \( h = 0 \) into the remaining limit:\[\lim_{h \to 0} \frac{1}{\sqrt{1+h}+1} = \frac{1}{\sqrt{1+0}+1} = \frac{1}{2}\]Thus, the limit evaluates to \( \frac{1}{2} \).

Key Concepts

Indeterminate Forms

Calculus often deals with limits, an essential concept for understanding the behavior of functions. When we approach a limit like \( \lim_{h \rightarrow 0} \frac{\sqrt{1+h}-1}{h} \), both the numerator and the denominator trend towards zero. This situation displays what's called an "indeterminate form," specifically \( \frac{0}{0} \).

Why is it indeterminate? When both parts of a fraction are zero, the expression doesn't provide a meaningful limit directly. It needs further manipulation to deduce a proper result. To recognize indeterminate forms:

• Look for expressions where both the numerator and denominator approach zero, like \( \frac{0}{0} \).
• Identify expressions that take an undefined form, like \( \frac{\infty}{\infty} \).

Indeterminate forms hint at the need for algebraic transformation or other calculus techniques to find the actual limit.

Rationalizing the Numerator

To manage the indeterminate form and solve the limit \( \lim_{h \rightarrow 0} \frac{\sqrt{1+h}-1}{h} \), we use a technique called "rationalizing the numerator."

Rationalizing involves multiplying the numerator and denominator by a conjugate, effectively transforming the expression. We choose the conjugate \( \sqrt{1+h}+1 \) for this task. This step is crucial for simplifying and removing the square root:

• Multiply numerator and denominator by \( \sqrt{1+h}+1 \).
• Transform \( \frac{\sqrt{1+h}-1}{h} \) into \( \frac{(\sqrt{1+h}-1)(\sqrt{1+h}+1)}{h(\sqrt{1+h}+1)} \).

This approach helps simplify complex expressions, often revealing a more straightforward path to calculate the limit.

Difference of Squares Formula

Using the "difference of squares" formula is a valuable step in simplifying the expression when rationalizing the numerator. The formula states:

• \((a-b)(a+b) = a^2 - b^2\)

In our expression, \(a\) is \(\sqrt{1+h}\) and \(b\) is 1, forming \((\sqrt{1+h}-1)(\sqrt{1+h}+1) = (1+h) - 1\). Applying the formula, this simplifies the numerator to \(h\), offering a much simpler expression \(\frac{h}{h(\sqrt{1+h}+1)}\).

Canceling \(h\) from numerator and denominator leaves us with \( \frac{1}{\sqrt{1+h}+1}\), which is no longer indeterminate because we can now directly substitute \(h = 0\) and compute the limit reliably as \( \frac{1}{2} \). Applying the difference of squares in calculus offers a technique to remove complexities introduced by square roots.