Problem 44 - Thomas Calculus in SI Units

Question

Use l'Hôpital's rule to find the limits. $$\lim _{h \rightarrow 0} \frac{e^{h}-(1+h)}{h^{2}}$$

Step-by-Step Solution

Answer

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

Step 1: Understand l'Hôpital's Rule

L'Hôpital's rule is used to solve limits of the form \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \). The rule states that if \( \lim_{x \to a} \frac{f(x)}{g(x)} = \frac{0}{0} \) or \( \frac{\infty}{\infty} \), then \( \lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a} \frac{f'(x)}{g'(x)} \), provided the latter limit exists.

Step 2: Identify the Limit Form

Evaluate the form of the limit \( \lim_{h \rightarrow 0} \frac{e^{h}-(1+h)}{h^{2}} \). Substituting \( h = 0 \) gives the indeterminate form \( \frac{0}{0} \), which means we can apply l'Hôpital's Rule.

Step 3: Differentiate the Numerator and Denominator

Find the derivative of the numerator \( e^h - (1 + h) \), which is \( f'(h) = e^h - 1 \). Find the derivative of the denominator \( h^2 \), which is \( g'(h) = 2h \).

Step 4: Apply L'Hôpital's Rule

Apply l'Hôpital's Rule: \( \lim_{h \rightarrow 0} \frac{e^{h}-(1+h)}{h^{2}} = \lim_{h \rightarrow 0} \frac{e^{h}-1}{2h} \). This limit is still in the form \( \frac{0}{0} \), so apply l'Hôpital's Rule again.

Step 5: Differentiate Again

Differentiate the numerator and the denominator again. The derivative of \( e^h - 1 \) is \( e^h \), and the derivative of \( 2h \) is \( 2 \).

Step 6: Evaluate the Second Limit

Evaluate the limit \( \lim_{h \rightarrow 0} \frac{e^h}{2} \). Substituting \( h = 0 \) gives \( \frac{e^0}{2} = \frac{1}{2} \).

Step 7: Conclusion

The original limit \( \lim_{h \rightarrow 0} \frac{e^{h}-(1+h)}{h^{2}} = \frac{1}{2} \). Thus, the solution using l'Hôpital's rule is complete.

Key Concepts

Understanding Limits in Calculus

In calculus, the concept of a limit is crucial for understanding how functions behave as their variables approach certain points. Limits help us examine points where functions are not explicitly defined or where direct substitution might not work. When we want to find the limit of a function as it approaches a particular value, we are essentially looking for the value that the function is trending towards as the input gets closer to a certain point. Take for example \( \lim_{h \rightarrow 0} \frac{e^{h}-(1+h)}{h^{2}} \). This expression investigates what happens to the function as \( h \) approaches zero.
If you substitute \( h = 0 \) directly, you might face a problem because the function tends to an indeterminate form. This is where limits become powerful, allowing us to compute values that aren't immediately apparent by evaluating the function directly. Limits are foundational for the rules and techniques used in calculus, including l'Hôpital's Rule.

The Nature of Indeterminate Forms

When working with limits, sometimes you encounter forms like \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \). These are known as indeterminate forms because they don't provide clear information about what the limit is. They tell us that more work is needed to resolve the limit. To navigate these challenges, we often employ strategies like algebraic manipulation or applying specific rules, such as l'Hôpital's Rule, to find a path to the solution.
In our example, substituting \( h = 0 \) into \( \lim_{h \rightarrow 0} \frac{e^{h}-(1+h)}{h^{2}} \) resulted in \( \frac{0}{0} \). This indicates an indeterminate form and signals the potential use of l'Hôpital's Rule. Being able to identify these forms is the first step in moving forward to finding the correct limit.

Solving Calculus Problems with L'Hôpital's Rule

L'Hôpital's Rule is a powerful tool in calculus for resolving indeterminate forms like \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \). It works by differentiating the numerator and the denominator separately, and then taking the limit again. In the exercise, applying l'Hôpital's Rule to \( \lim_{h \rightarrow 0} \frac{e^{h}-(1+h)}{h^{2}} \) involves taking the derivative of the numerator to get \( e^h - 1 \) and the derivative of the denominator to get \( 2h \).
This simplifies our limit to \( \lim_{h \rightarrow 0} \frac{e^h - 1}{2h} \), which is still an indeterminate form. Applying l'Hôpital's Rule once more, the derivatives become \( e^h \) for the numerator and \( 2 \) for the denominator. The limit becomes much simpler: \( \lim_{h \rightarrow 0} \frac{e^h}{2} \). When evaluated, this final limit is \( \frac{1}{2} \). Understanding and applying l'Hôpital's Rule is a key aspect of mastering calculus problem-solving, providing clarity in situations that initially seem unresolved.