Problem 19 - Calculus for Biology and Medicine

Question

Find the limits in Problems 1-60; not all limits require use of l'Hôpital's rule. $$ \lim _{x \rightarrow \infty} \frac{e^{x}-1-x}{e^{x}-x^{2}} $$

Step-by-Step Solution

Answer

The limit is 1.

Step 1: Recognize the form

Examine the expression to recognize the form. The limit is \(\lim_{x \rightarrow \infty} \frac{e^x - 1 - x}{e^x - x^2}\). As \(x\) approaches infinity, both the numerator \(e^x - 1 - x\) and the denominator \(e^x - x^2\) approach infinity, indicating an indeterminate form \(\frac{\infty}{\infty}\). Thus, l'Hôpital's rule can be applied.

Step 2: Apply l'Hôpital's Rule

Apply l'Hôpital's Rule, which states that for limits of the form \(\frac{\infty}{\infty}\) or \(\frac{0}{0}\), the limit of a quotient is the limit of the quotient of their derivatives. Differentiate the numerator: \(\frac{d}{dx}[e^x - 1 - x] = e^x - 1\). Differentiate the denominator: \(\frac{d}{dx}[e^x - x^2] = e^x - 2x\). The limit becomes \(\lim_{x \rightarrow \infty} \frac{e^x - 1}{e^x - 2x}\).

Step 3: Simplify and Re-evaluate the Limit

Evaluate the limit of \(\frac{e^x - 1}{e^x - 2x}\). Divide both the numerator and the denominator by \(e^x\): \(\lim_{x \rightarrow \infty} \frac{1 - e^{-x}}{1 - \frac{2x}{e^x}}\). As \(x \rightarrow \infty\), \(e^{-x} \rightarrow 0\) and \(\frac{2x}{e^x} \rightarrow 0\), simplifying the expression to \(\frac{1-0}{1-0} = 1\).

Step 4: Conclusion

Since further simplification remains indeterminate, apply l'Hôpital's rule again if necessary. However, in this case, the limit resolves to a constant upon the first application of L'Hôpital's Rule, and therefore further applications are unnecessary. Concluding with the found result of the limit.

Key Concepts

Indeterminate Forms

In calculus, indeterminate forms are situations where a limit does not initially convey a clear, direct value. Such conditions occur primarily with expressions like \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\). These forms are called "indeterminate" because they require further manipulation to decide what value a limit approaches.

When faced with a limit that leads to an indeterminate form, it indicates that more analysis is required. This often involves methods like simplifying the expression, employing algebraic manipulations, or leveraging specific theorems or tools such as l'Hôpital's Rule.

For instance, in the exercise provided, the expression \(\frac{e^x - 1 - x}{e^x - x^2}\) as \(x\) approaches infinity, leads to the \(\frac{\infty}{\infty}\) form. This indeterminate form signals that neither the numerator nor the denominator restricts the value directly. Thus, l'Hôpital's Rule proves useful to find the limit.

Limits in Calculus

Limits form the backbone of calculus. They describe the behavior of functions as they get close to a certain point, yet never quite reach it. Understanding limits is essential for working with continuity, derivatives, and integrals.

Limits answer questions like, "What does \(f(x)\) approach as \(x\) moves nearer to a particular point?" In the context of the exercise, we're interested in finding the limit of the function \(\frac{e^x - 1 - x}{e^x - x^2}\) as \(x\) heads towards infinity. Since direct substitution results in an indeterminate form, we need to work through algebraic or rule-based simplification to find the limit.

In calculus, limits often serve as a machine to "zoom in" on function behavior at crucial points. When dealing with sequences, derivatives, or integrals, knowing how to handle limits equips one with a deeper understanding of continuous change.

Exponential Functions

Exponential functions, notably \(e^x\), play a significant role in both mathematical theory and practical applications. The exponential function \(y = e^x\) features prominently because it grows very rapidly as \(x\) increases.

One of its extraordinary properties is that the derivative of \(e^x\) is itself \(e^x\), making it essential in differential calculus. Because of this property, exponential functions frequently appear in growth and decay models, finance calculations, and many areas of science.

In the problem provided, \(e^x\) dominates both the numerator and the denominator due to its rapid growth rate, which is why dividing each part by \(e^x\) significantly simplifies the expression in the limit analysis process. Understanding exponential functions and their behavior is critical when manipulating limits involving them, as they often influence the outcome due to their powerful presence.