Question

Evaluate the following limits. $$\lim _{x \rightarrow 0} \frac{a^{x}-b^{x}}{x}, \text { for positive constants } a \text { and } b$$

Short answer

Question: Evaluate the following limit: $$\lim _{x \rightarrow 0} \frac{a^{x}-b^{x}}{x}$$ Answer: The limit of the given function as x approaches 0 is: $$\lim _{x \rightarrow 0} \frac{a^{x}-b^{x}}{x} = \ln \left(\frac{a}{b}\right)$$

Step-by-step solution

Identify the indeterminate form

The given limit is of the form \(\frac{0}{0}\), as when we plug in x = 0, we get \(\frac{a^0 - b^0}{0} = \frac{1-1}{0}\). Since this is an indeterminate form, we will proceed by applying L'Hopital's rule.

Differentiate the numerator and the denominator with respect to x

To apply L'Hopital's rule, we need to find the derivatives of the numerator and the denominator with respect to x. So, we have: \(\text{Numerator: } \frac{d}{dx}(a^x - b^x) = a^x \ln a - b^x \ln b\) \(\text{Denominator: } \frac{d}{dx} (x) = 1\)

Apply L'Hopital's rule

Now, we will apply L'Hopital's rule, which states that if the limit of the ratio of the derivatives exists, then the limit of the original function is equal to the limit of the ratio of the derivatives. Thus, the limit we want to find is: $$\lim_{x \rightarrow 0} \frac{(a^x \ln a - b^x \ln b)}{1}$$

Evaluate the limit

Plugging x = 0 into the expression gives: $$\frac{(a^0 \ln a - b^0 \ln b)}{1} = \frac{(\ln a - \ln b)}{1}$$ As we have a subtraction of natural logarithms, we can simplify the result: \(\frac{(\ln a - \ln b)}{1} = \frac{\ln \left(\frac{a}{b}\right)}{1} = \ln \left(\frac{a}{b}\right)\) Therefore, the limit of the given function as x approaches 0 is: $$\lim _{x \rightarrow 0} \frac{a^{x}-b^{x}}{x} = \ln \left(\frac{a}{b}\right)$$

Key concepts

L'Hopital's Rule

In calculus, L'Hopital's rule is a valuable tool for evaluating limits that results in indeterminate forms like \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \) by deriving the functions in the numerator and the denominator. This rule is particularly useful because it transforms challenging limits into simpler ones that we can evaluate directly.

To apply L'Hopital's rule, start by differentiating the top and bottom of the limit separately. If after taking the derivatives, the limit is no longer indeterminate, you can usually just plug in the value of x to get the answer. Sometimes, applying L'Hopital's rule once isn't enough, and you might have to repeat it until you obtain a determinate form. Always make sure the conditions for L'Hopital's rule are met before applying it, which include the original limit being indeterminate and the derivatives of numerator and denominator existing around the point of interest.

Indeterminate Forms

Indeterminate forms arise in calculus when we try to evaluate certain limits, and the standard techniques, like direct substitution, don't yield a definite answer. These forms include \( \frac{0}{0} \), \( \frac{\infty}{\infty} \), \( 0 \cdot \infty \), \( \infty - \infty \), \( 0^0 \), \( \infty^0 \), and \( 1^\infty \). They signal that further work is required to find the limit and that other approaches, such as L'Hopital's rule, factorization, or conjugation, might be needed.

The indeterminate form \( \frac{0}{0} \) implies that both the numerator and denominator approach zero, making the limit's behavior uncertain without additional information. Determining the actual value involves algebraic manipulation or methods like L'Hopital's rule to resolve the indeterminacy and compute the limit.

Derivatives of Exponential Functions

When dealing with limits involving exponential functions, knowing how to differentiate these functions is crucial. The derivative of an exponential function \( a^x \) with respect to x is given by \( a^x \ln(a) \) if a is a positive constant and not equal to 1.

This particular derivative allows us to apply L'Hopital's rule to exponential functions in indeterminate form. The logarithmic factor, \( \ln(a) \) in the derivative, comes from the chain rule, because the base a is raised to a power that depends on x. Understanding this concept is fundamental when evaluating limits of functions involving exponentials, as it uncovers the rate at which these functions grow or decrease at a particular point, and it is especially valuable in solving limits similar to the exercise provided.