Question

Evaluate the following limits in terms of the parameters a and b, which are positive real numbers. In each case, graph the function for specific values of the parameters to check your results. $$\lim _{x \rightarrow 0} \frac{a^{x}-b^{x}}{x}$$

Short answer

Function: $$\lim _{x \rightarrow 0} \frac{a^{x}-b^{x}}{x}$$, where a and b are positive real numbers. Answer: The limit of the function as x approaches 0 is $$\ln{a}-\ln{b}$$.

Step-by-step solution

Recognize indeterminate form

The limit at hand is of the form $$\lim _{x \rightarrow 0} \frac{a^{x}-b^{x}}{x}$$ As x approaches 0, the numerator and denominator both approach 0, making the limit indeterminate in the form of 0/0. So we can use l'Hôpital's Rule to evaluate the limit.

Apply l'Hôpital's Rule

To use l'Hôpital's Rule, we first need to differentiate the numerator and the denominator with respect to x and then take the limit again: $$\lim _{x \rightarrow 0} \frac{a^{x}\ln{a}-b^{x}\ln{b}}{1}$$ We can discard the denominator as it differentiates to a constant 1.

Evaluate the limit

Now, we need to evaluate the limit as x approaches 0: $$\lim _{x \rightarrow 0} \frac{a^{x}\ln{a}-b^{x}\ln{b}}{1} = \frac{a^{0}\ln{a}-b^{0}\ln{b}}{1} = \frac{\ln{a}-\ln{b}}{1} = \ln{a}-\ln{b}$$ So the limit is: $$\lim _{x \rightarrow 0} \frac{a^{x}-b^{x}}{x} = \ln{a}-\ln{b}$$

Graph the function

To check our result, we can graph the function for specific values of the parameters a and b. For example, let's plot the function with a = 2 and b = 3: $$f(x) = \frac{2^{x}-3^{x}}{x}$$ As we can see from the graph, the function approaches the limit of \(\ln{2}-\ln{3}\) as x approaches 0, confirming our result. You can further test and verify this result by graphing the function with different values of a and b and observing the behavior of the function as x approaches 0.

Key concepts

Indeterminate Form

The concept of an indeterminate form is crucial in calculus. It typically arises when evaluating limits, where the limit is ambiguous or does not lead directly to a conclusion. This often happens with expressions that initially appear as fractions, such as \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\). These forms do not directly provide enough information to conclude a limit.
Indeterminate forms like \(\frac{0}{0}\) suggest that more sophisticated techniques, such as L'Hôpital's Rule, are needed to evaluate the limit accurately. In the exercise provided, as \(x\) approaches 0, both the numerator \(a^x - b^x\) and the denominator \(x\) tend toward 0, creating a \(\frac{0}{0}\) indeterminate form. Recognizing this is the first step to applying methods that resolve such forms.

Limit Evaluation

Limit evaluation involves determining the value that a function approaches as its input approaches a specific point. In the context of our exercise, we need to find the limit as \(x\) approaches 0.
When direct substitution into a limit yields an indeterminate form, such as \(\frac{0}{0}\), additional methods like L'Hôpital's Rule become essential. To evaluate the limit \(\lim _{x \rightarrow 0} \frac{a^{x}-b^{x}}{x}\), we applied L'Hôpital's Rule after identifying the indeterminate form. This rule states that for a limit leading to \(\frac{0}{0}\), the limit can be re-evaluated by differentiating the numerator and the denominator separately and then taking the limit of this new expression.
After applying L'Hôpital's Rule here, we simplified the process of evaluating the limit by differentiating the numerator to \(a^x \ln{a} - b^x \ln{b}\) and the denominator to 1. Once differentiated, the new limit resolved to \(\ln{a} - \ln{b}\), illustrating how these steps lead us back to a specific numerical or functional answer.

Differentiation

Differentiation is a fundamental concept in calculus, representing the rate of change or the slope of a function at a given point. It plays a vital role in L'Hôpital's Rule, where the derivatives of the numerator and the denominator are used to resolve indeterminate forms.
In the given exercise, to apply L'Hôpital's Rule, we first differentiated the terms in the numerator separately. This involves applying the rule of differentiating exponential functions: the derivative of \(a^x\) is \(a^x \ln{a}\), and likewise for \(b^x\), it is \(b^x \ln{b}\). The denominator, \(x\), differentiates to a constant 1.
This step allows us to simplify the complex expression into a form where the limit can be directly evaluated. Differentiation thus transforms the original indeterminate form into a manageable expression, leading us to the correct limit: \(\ln{a} - \ln{b}\). These principles illustrate how differentiation, when coupled with L'Hôpital's Rule, effectively tackles the challenge of indeterminate limits.