Problem 44 - Thomas Calculus

Question

Use l'Hopital's rule to find the limits in Exercises \(7-50\) . $$ \lim _{h \rightarrow 0} \frac{e^{h-(1+h)}}{h^{2}} $$

Step-by-Step Solution

Answer

The limit is 0.

Step 1: Identify the Form

We need to determine if the limit \( \lim _{h \rightarrow 0} \frac{e^{h-(1+h)}}{h^{2}} \) is an \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \) indeterminate form. When \( h = 0 \), the numerator becomes \( e^{0} = 1 \) and the denominator is \( 0^2 = 0 \), so the expression becomes \( \frac{1-1}{0} = \frac{0}{0} \). This is an indeterminate form suitable for l'Hopital's Rule.

Step 2: Apply l'Hopital's Rule

Since the form is \( \frac{0}{0} \), we can apply l'Hopital's Rule. This rule states that \( \lim _{x \to a} \frac{f(x)}{g(x)} = \lim _{x \to a} \frac{f'(x)}{g'(x)} \) if the limit exists. First, we find the derivatives of the numerator and denominator.

Step 3: Differentiate the Numerator

The numerator is \( e^{h-(1+h)} = e^{h-h-1} = e^{-1} \). Differentiating with respect to \( h \), we get \( \frac{d}{dh}(e^{-1}) = 0 \). This is because \( e^{-1} \) is a constant.

Step 4: Differentiate the Denominator

The denominator is \( h^2 \). Differentiating with respect to \( h \), we get \( 2h \).

Step 5: Compute the New Limit

Using the derivatives found in Steps 3 and 4, apply l'Hopital's Rule: \[\lim _{h \to 0} \frac{0}{2h} = \lim _{h \to 0} 0 = 0.\]

Step 6: Conclusion

Since the application of l'Hopital's Rule resulted in \( 0 \), the limit of the original expression as \( h \) approaches 0 is 0.

Key Concepts

Calculus Limits

Calculus limits are a fundamental concept in calculus, representing the value that a function approaches as the input approaches some point. Calculating limits helps us understand the behavior of functions at points that might not be clearly defined in the function itself.
In the given exercise, the limit we are asked to evaluate is:

• The expression \( \lim _{h \rightarrow 0} \frac{e^{h-(1+h)}}{h^{2}} \).

When approaching limits, especially when the variable approaches zero, it's important to meticulously assess the components of the function involved. In particular, the exponential function's continuity plays a vital role, combined with polynomial expressions in the denominator.
This initial step of determining the form of a limit sets the stage for the techniques that we will use to solve it effectively.

Indeterminate Forms

Indeterminate forms are expressions in calculus that do not initially present a clear path to evaluation. The most common indeterminate forms are \( \frac{0}{0} \) and \( \frac{\infty}{\infty} \), but they can take various other forms as well, like \( \infty - \infty \) or \( 0 \times \infty \).
In the problem at hand, by substituting \( h = 0 \), the resulting form is \( \frac{0}{0} \), which is an indeterminate form.

• This means direct substitution is not possible to find the limit.
• l'Hopital's Rule provides a pathway to evaluate such limits by transforming them using derivatives.
• This transformation allows for evaluating something that initially seems unsolvable by conventional limit tactics.

Identifying indeterminate forms is crucial because they signal the need for special limit techniques, such as l'Hopital's Rule.

Derivatives

Derivatives express how a function changes as its input changes and are central to many calculus concepts, including limit evaluation. In the context of l'Hopital's Rule, derivatives transform indeterminate forms into more manageable expressions.
In this exercise, we determined:

• The numerator \( e^{-1} \) in the expression is constant, so its derivative is \( 0 \).
• The polynomial \( h^2 \) has a derivative of \( 2h \).

Finding these derivatives is key to applying l'Hopital's Rule correctly. These derivatives essentially allow the application of the rule by breaking down each component separately, turning the originally unmanageable indeterminate form into a simpler expression.

Limit Evaluation Techniques

Limit evaluation techniques in calculus are methods used to compute limits, especially when direct substitution isn't possible or doesn't initially make sense.
An important technique illustrated in this problem is l'Hopital's Rule, which is applicable for certain indeterminate forms. The rule dictates that:

• If we have a \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \) form, we take the derivative of the numerator and denominator separately.
• The new limit of the derivatives is the same as the original expression if the limit of the derivatives exists.

In this case, after applying l'Hopital's Rule, the new expression becomes \( \frac{0}{2h} \), leading to a straightforward computation of the limit, which is \( 0 \). This demonstrates the power and utility of l'Hopital's Rule in simplifying complex indeterminate forms into solvable problems.