Question

Limits Evaluate the following limits. Use l'Hópital's Rule when it is comvenient and applicable. $$\lim _{x \rightarrow \infty} \frac{e^{1 / x}-1}{1 / x}$$

Short answer

Question: Find the value of the limit: $$\lim _{x \rightarrow \infty} \frac{e^{1 / x}-1}{1 / x}$$ Answer: The value of the given limit is 1.

Step-by-step solution

Identify the given limit

We are given the following limit to evaluate: $$\lim _{x \rightarrow \infty} \frac{e^{1 / x}-1}{1 / x}$$

Check if l'Hôpital's Rule is applicable

As x approaches infinity, the given limit will be in the form of \(\frac{0}{0}\), since both the numerator and denominator will approach zero. As this is an indeterminate form, we can apply l'Hôpital's Rule.

Differentiate the numerator and denominator

Using the chain rule, calculate the derivatives of the numerator and denominator: Numerator: \(y = e^{1 / x} - 1\) \(\frac{dy}{dx} = -\frac{1}{x^2} e^{1 / x}\), as the derivative of a constant (1) is 0. Denominator: \(y = \frac{1}{x}\) \(\frac{dy}{dx} = -\frac{1}{x^2}\)

Apply l'Hôpital's Rule and evaluate the limit

According to l'Hôpital's Rule: $$\lim _{x \rightarrow \infty} \frac{e^{1 / x}-1}{1 / x} = \lim _{x \rightarrow \infty} \frac{-\frac{1}{x^2} e^{1 / x}}{-\frac{1}{x^2}}$$ Simplify the expression: $$\lim _{x \rightarrow \infty} e^{1 / x}$$ Now, evaluating the limit as x approaches infinity, we get: $$\lim _{x \rightarrow \infty} e^{1 / x} = e^{1 / \infty} = e^{0} = 1$$ Therefore, the given limit is: $$\lim _{x \rightarrow \infty} \frac{e^{1 / x}-1}{1 / x} = 1$$

Key concepts

Indeterminate Forms

When working with limits in calculus, you may encounter expressions that are not immediately clear or definable. These are known as indeterminate forms. They typically arise when you have competing tendencies that result in a form like \(0/0\), \(\infty/\infty\), \(0 \cdot \infty\), \(\infty-\infty\), or similar. For instance, when evaluating the limit \(\lim _{x \rightarrow \infty} \frac{e^{1 / x}-1}{1 / x}\), as x grows larger, both the numerator and the denominator approach zero, yielding an indeterminate form of \(0/0\). This is where tools like L'Hôpital's Rule can be invaluable, as they provide a means to evaluate these limits.

L'Hôpital's Rule specifically deals with these indeterminate forms by allowing us to replace the initial limit with a potentially easier to evaluate limit of derivatives. Ensuring the conditions are met — both the numerator and denominator must approach zero or infinity — is the first step when applying L'Hôpital's Rule.

Derivatives in Calculus

The derivatives of functions in calculus are fundamental when analyzing change. In the context of evaluating limits using L'Hôpital's Rule, finding the derivative is a necessary step, as we replace the original indeterminate form with the limit of the derivatives of the numerator and denominator. In our exercise example, differentiating \(e^{1 / x} - 1\) and \(1/x\) is necessary to apply L'Hôpital's Rule.

The derivative of \(e^{1 / x}\) involves an understanding of the exponential function and the chain rule. Proper differentiation is vital here as incorrect derivatives will lead to wrong limit results. Mastery of derivatives also equips students with the ability to tackle a wide array of problems in calculus beyond limits, including optimization and modeling motion.

Chain Rule

The chain rule is a powerful derivative rule in calculus, which is used to find the derivative of the composition of two or more functions. For the derivative of \(e^{1 / x}\), we first recognize that \(1/x\) is a function within the exponential function. Using the chain rule, we differentiate the outer function (the exponential) at the inner function (the reciprocal of x) and multiply it by the derivative of the inner function.

In our step by step solution, the chain rule allows us to find the derivative \(\frac{dy}{dx} = -\frac{1}{x^2} e^{1 / x}\), effectively handling the composite function. Learning and applying the chain rule is crucial, not just for using L'Hôpital's Rule, but for all instances where you encounter nested functions in calculus.

Exponential Functions

Exponential functions, generally expressed in the form \(e^x\) or \(a^x\), where e is Euler's number (approximately 2.718) and a is any positive constant, are pivotal in calculus for depicting growth or decay processes. They possess the unique property that the function is its own derivative, which simplifies many calculus problems.

In the given exercise, we deal with the exponential function \(e^{1 / x}\). As x approaches infinity, \(1/x\) approaches zero, and thus \(e^{1 / x}\) approaches \(e^0\), which is 1. Understanding the behavior of exponential functions not only aids in evaluating limits, but also in the broader contexts within calculus and applied mathematics, such as in the study of compound interest, population growth, and radioactive decay.