Problem 44 - University Calculus: Early Transcendentals

Question

Use I'Hópital'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: Identify the Indeterminate Form

First, we substitute the limit value directly into the expression: \[\lim _{h \rightarrow 0} \frac{e^{h}-(1+h)}{h^{2}}\]This gives us \[\frac{e^{0}-(1+0)}{0^2} = \frac{1-1}{0} = \frac{0}{0},\]which is an indeterminate form. Therefore, we can apply L'Hôpital's rule.

Step 2: Apply L'Hôpital's Rule

Given the form \(\frac{0}{0}\), apply L'Hôpital's Rule which involves differentiating the numerator and the denominator.The derivative of the numerator, \(e^h - (1 + h)\), is \(e^h - 1\).The derivative of the denominator, \(h^2\), is \(2h\).Thus, applying L'Hôpital's Rule:\[\lim _{h \rightarrow 0} \frac{e^{h} - 1}{2h}\]

Step 3: Check the New Indeterminate Form

Substitute \(h = 0\) into the new expression:\[\frac{e^{0} - 1}{2(0)} = \frac{0}{0},\]an indeterminate form again. So, we apply L'Hôpital's Rule once more.

Step 4: Apply L'Hôpital's Rule Again

Differentiate the numerator, \(e^h - 1\), to get \(e^h\).Differentiate the denominator, \(2h\), to get \(2\).Now apply L'Hôpital's Rule a second time:\[\lim _{h \rightarrow 0} \frac{e^{h}}{2}.\]

Step 5: Evaluate the Limit

Substitute \(h = 0\) into the expression:\[\lim _{h \rightarrow 0} \frac{e^{h}}{2} = \frac{e^{0}}{2} = \frac{1}{2}.\]Therefore, the limit is \(\frac{1}{2}\).

Key Concepts

Indeterminate Forms

In calculus, indeterminate forms arise when evaluating limits, where direct substitution results in ambiguous expressions such as \( \frac{0}{0} \). These forms do not give immediate information about the limit, making the calculation tricky. The expression \( \lim_{h \rightarrow 0} \frac{e^{h}-(1+h)}{h^{2}} \) initially resolves into one of these expressions: \( \frac{0}{0} \). Recognizing this is crucial because it suggests the necessity for tools like L'Hôpital's Rule.

• Types of Indeterminate Forms: Besides \( \frac{0}{0} \), there are others like \( \frac{\infty}{\infty} \), \( 0 \times \infty \), \( \infty - \infty \), \( 0^0 \), \( \infty^0 \), and \( 1^\infty \).
• Application: When you encounter these forms, it indicates that further manipulation of the expression or application of a special rule like L'Hôpital’s is necessary to evaluate the limit.

Identifying an indeterminate form is the first step toward navigating through challenging limit problems.

Calculus Limits

Limits are fundamental in calculus, serving as a foundation for defining derivatives and integrals. They describe the behavior of functions as inputs approach a certain point. In the problem \( \lim_{h \rightarrow 0} \frac{e^{h}-(1+h)}{h^{2}} \), we're interested in the value the function approaches as \( h \) gets infinitely close to zero. This example demonstrates typical limit evaluation steps, including substitution and simplification.

• Direct Substitution: The initial step where you plug in the limit directly to see if it results in a real number or an indeterminate form.
• Simplification Techniques: Applying algebraic manipulations or calculus rules (like L'Hôpital's) when direct substitution fails as it leads to \( \frac{0}{0} \) or similar forms.
• Defining Continuity and Derivatives: Limits help in describing the smoothness of functions and calculating rates of change.

Understanding limits is not only essential for evaluating specific problems but for grasping the robust framework of calculus.

Differentiation

Differentiation is a core calculus operation used to compute the rate at which functions change. In solving limits via L'Hôpital's Rule, differentiation plays a pivotal role. The rule involves taking derivatives of the numerator and denominator independently to simplify indeterminate forms. For instance, in the exercise \( \lim_{h \rightarrow 0} \frac{e^{h}-(1+h)}{h^{2}} \), differentiation of the numerator and denominator leads to solvable expressions.

• Power of Differentiation: It allows the transformation of the original indeterminate into a simpler limit by questioning how the function changes as inputs vary.
• Higher-order Differentiation: Sometimes, repeated differentiation is needed to resolve complex limits, as seen with the second application of L'Hôpital's Rule here.
• Practical Applications: Beyond limits, differentiation helps in finding slopes of curves, optimizing problems, and modeling scientific phenomena mathematically.

Mastering differentiation is invaluable for tackling calculus problems, especially when tools like L'Hôpital's Rule are required.