Problem 15 - Thomas Calculus

Question

Find the limits in Exercises \(11-18\) $$\lim _{h \rightarrow 0^{+}} \frac{\sqrt{h^{2}+4 h+5}-\sqrt{5}}{h}$$

Step-by-Step Solution

Answer

The limit is \( \frac{2 \sqrt{5}}{5} \).

Step 1: Recognize the Indeterminate Form

First, notice that as \( h \rightarrow 0^{+} \), the expression \( \frac{\sqrt{h^{2}+4h+5} - \sqrt{5}}{h} \) becomes indeterminate (\( \frac{0}{0} \) form). This indicates that algebraic manipulation or other techniques are needed to find the limit.

Step 2: Rationalize the Numerator

To handle the indeterminate form, rationalize the numerator by multiplying the expression by \( \frac{\sqrt{h^{2}+4h+5} + \sqrt{5}}{\sqrt{h^{2}+4h+5} + \sqrt{5}} \). This results in:\[\frac{h^2 + 4h + 5 - 5}{h(\sqrt{h^{2}+4h+5} + \sqrt{5})}\]which simplifies to:\[\frac{h(h + 4)}{h(\sqrt{h^{2}+4h+5} + \sqrt{5})}\]

Step 3: Simplify the Expression

Cancel out the \( h \) in the numerator and the denominator, as long as \( h eq 0 \):\[\frac{h + 4}{\sqrt{h^{2}+4h+5} + \sqrt{5}}\]

Step 4: Evaluate the Limit as \( h \rightarrow 0^{+} \)

Now, substitute \( h = 0 \) into the expression:\[\frac{0 + 4}{\sqrt{0^{2}+4 \times 0 + 5} + \sqrt{5}} = \frac{4}{\sqrt{5} + \sqrt{5}} = \frac{4}{2\sqrt{5}}\]Thus, simplifying gives:\[\frac{2}{\sqrt{5}} = \frac{2 \sqrt{5}}{5}\]

Step 5: Final Answer

The limit evaluates to \( \frac{2 \sqrt{5}}{5} \), which is the final answer.

Key Concepts

Indeterminate Forms

When we're working with limits, especially as values approach zero, we often encounter expressions that result in an indeterminate form. One of the most common of these forms is the \( \frac{0}{0} \) form, which means both the numerator and denominator approach zero.
Indeterminate forms are problematic because they don't provide enough information to determine the limit outright. Instead, they suggest we need to use additional methods to better understand the behavior of the function near the point of interest.

• In the original exercise, we see that as \( h \to 0^+ \), substituting directly would result in the expression \( \frac{\sqrt{5}-\sqrt{5}}{0} \), or \( \frac{0}{0} \).
• This indicates a need for algebraic manipulation like rationalization, differentiation, or other limit techniques to resolve this form.

Recognizing indeterminate forms is crucial because once identified, we know it's not the final answer but a step in finding it.

Rationalizing Technique

When faced with an indeterminate form involving square roots, the rationalizing technique becomes essential. Rationalizing is the process of eliminating the radicals in the denominator or numerator to simplify the expression.
In this context, we multiply by the conjugate, which involves changing the sign between two terms that form a difference. This helps eliminate the square root in the numerator.

• From our exercise, the conjugate of \( \sqrt{h^2 + 4h + 5} - \sqrt{5} \) is \( \sqrt{h^2 + 4h + 5} + \sqrt{5} \).
• By multiplying the fraction by \( \frac{\sqrt{h^2 + 4h + 5} + \sqrt{5}}{\sqrt{h^2 + 4h + 5} + \sqrt{5}} \), we simplify the numerator into \( h(h+4) \).

This reduces complexity and helps cancel terms that otherwise prevent further simplification. It's a vital technique for handling indeterminate solutions with roots.

Evaluating Limits

Once rationalization addresses the indeterminate form, the next step is evaluating the limit of the simplified expression. Evaluating limits determines how a function behaves as the input approaches a certain value, which is a fundamental concept in calculus.
With the simplified form \( \frac{h+4}{\sqrt{h^2 + 4h + 5} + \sqrt{5}} \), we can now substitute \( h = 0 \) with confidence.

• Upon substitution, the expression reduces to \( \frac{4}{2\sqrt{5}} \), leading to a final result of \( \frac{2\sqrt{5}}{5} \).
• This step verifies that there are no remaining indeterminacies or undefined operations.

Evaluating limits with methods like substitution post-simplification ensures accuracy and bridges gaps caused by initial indeterminate forms.