Problem 19 - Calculus for Biology and Medicine

Question

Use l'Hospital's rule to find the limits. $$ \lim _{x \rightarrow 0} \frac{e^{x}-1-x}{x^{2}} $$

Step-by-Step Solution

Answer

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

Step 1: Identify the indeterminate form

First, substitute the value of the limit into the function: \( \frac{e^x - 1 - x}{x^2} \). This results in \( \frac{e^0 - 1 - 0}{0^2} = \frac{0}{0} \), which is an indeterminate form. This means we can apply l'Hospital's rule.

Step 2: Apply l'Hospital's Rule

Since the limit is in the \(\frac{0}{0}\) indeterminate form, we can apply l'Hospital's Rule, which states that \( \lim_{x \to c} \frac{f(x)}{g(x)} = \lim_{x \to c} \frac{f'(x)}{g'(x)} \) provided the latter limit exists. Here, we differentiate the numerator and the denominator separately.

Step 3: Differentiate the numerator and the denominator

Differentiate the numerator \( e^x - 1 - x \) with respect to \( x \) to get \( e^x - 1 \). Differentiate the denominator \( x^2 \) with respect to \( x \) to get \( 2x \).

Step 4: Substitute and simplify after differentiation

The new expression after applying l'Hospital's Rule is \( \frac{e^x - 1}{2x} \). Substitute \( x = 0 \) to check if this yields a determinate form. We again find \( \frac{e^0 - 1}{2 \times 0} = \frac{0}{0} \), another indeterminate form. This means we need to apply l'Hospital's Rule again.

Step 5: Apply l'Hospital's Rule again

Differentiate the new numerator \( e^x - 1 \) to get \( e^x \). Differentiate the new denominator \( 2x \) to get \( 2 \).

Step 6: Evaluate the simplified limit

The expression is simplified to \( \frac{e^x}{2} \). Substituting \( x = 0 \) now gives us \( \frac{e^0}{2} = \frac{1}{2} \). This is no longer an indeterminate form.

Key Concepts

Indeterminate Forms

Indeterminate forms occur in calculus when calculating a limit directly results in expressions like \(\frac{0}{0}\), \(\frac{\infty}{\infty}\), \(0 \times \infty\), and more. These forms do not immediately provide useful information about the limit, making them a primary target for further analysis.

• For example, a function approaching \(\frac{0}{0}\) suggests that both the numerator and denominator approach zero as the variable gets close to a certain point, but it is unclear what the quotient approaches without further breakdown.
• These instances often imply that the function's behavior is more complex, and more tools, such as l'Hospital's Rule, are needed to evaluate the limit.

By identifying the indeterminate form in the original problem, like \(\frac{0}{0}\) as seen here, we recognize why direct substitution does not work and know that a deeper examination is required.

Calculus Limits

The concept of limits is fundamental in calculus. Limits describe the behavior of a function as the input approaches a specific value. Understanding limits allows mathematicians to handle continuous functions, curves, and rates of change mathematically.

• The limit, denoted \(\lim_{x \to c} f(x)\), tells us what value \(f(x)\) is approaching as \(x\) gets closer to \(c\).
• Calculating limits directly can be simple, but when indeterminate forms arise, it becomes complex, often requiring analytical techniques such as l'Hospital's Rule.
• For students, mastering limits can lead to a deeper understanding of other calculus operations, as they underpin differentiation and integration.

In problems like the one we're examining, identifying when to use tools like limits and knowing their properties is crucial in peeling back the complexity of challenging functions.

Differentiation

Differentiation is the process through which we calculate the derivative of a function. It is a core tool in calculus used to understand rates of change, slopes, and curve behaviors.

• Equation of differentiation: If \( f(x) \) is a function, its derivative \( f'(x) \) represents the function's rate of change at any given point \(x\).
• In the context of l'Hospital's Rule, differentiation helps "simplify" indeterminate forms. By differentiating both the numerator and the denominator of a fraction, we attempt to find a new limit that exists in a determinate form.
• For instance, in this exercise, successive differentiation led us to an expression \(\frac{e^x}{2}\), enabling evaluation of the limit to \(\frac{1}{2}\) upon substitution.

By understanding differentiation, students can tackle more advanced problems in calculus, particularly those involving troublesome limits and unexpected forms.