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. $$\lim _{x \rightarrow 0^{+}}\left(a^{x}-b^{x}\right)^{1 / x}, a>b>0$$

Short answer

Based on the step-by-step solution provided, answer the following question: Question: Evaluate the limit of the given function as x approaches 0 from the positive side: $$ \lim _{x \rightarrow 0^{+}}\left(a^{x}-b^{x}\right)^{1 / x} $$ where a > b > 0. Answer: The limit of the given function as x approaches 0 from the positive side is: $$ \frac{a}{b} $$

Step-by-step solution

Rewrite the expression using natural logarithm property

To address the problem of the expression having a fractional exponent, rewrite the expression in terms of natural logarithms using the following property: $$\ln (a^x) = x \ln a$$ The given expression can then be written as: $$ e^{\left(\lim _{x \rightarrow 0^{+}}\frac{\ln(a^x)-\ln(b^x)}{x}\right)} $$

Apply L'Hopital's Rule

Since we have a 0/0 indeterminate form, L'Hopital's Rule can be applied. We need to take derivatives of both the numerator and the denominator with respect to x. Numerator: $$\frac{d}{dx}(\ln(a^x)-\ln(b^x)) = \frac{d}{dx}(x \ln a - x \ln b) = \ln a-\ln b$$ Denominator: $$\frac{d}{dx}(x) = 1$$ After applying L'Hopital's Rule, we get: $$ \lim _{x \rightarrow 0^{+}}\frac{\ln(a^x)-\ln(b^x)}{x} = \lim _{x \rightarrow 0^{+}}(\ln a - \ln b) = \ln a - \ln b $$

Substitute the limit back into the original expression

Now that we have evaluated the limit of the exponent, we can substitute it back into the original expression: $$ e^{\left(\lim _{x \rightarrow 0^{+}}\frac{\ln(a^x)-\ln(b^x)}{x}\right)} = e^{\ln a - \ln b} $$

Simplify the final expression

Using the property of logarithms, we can express the difference of logs as a quotient: $$ e^{\ln a - \ln b} = e^{\ln \left(\frac{a}{b}\right)} $$ Then, using the property of exponentials, we have: $$ e^{\ln \left(\frac{a}{b}\right)} = \frac{a}{b} $$ Therefore, the limit is: $$ \lim _{x \rightarrow 0^{+}}\left(a^{x}-b^{x}\right)^{1 / x} = \frac{a}{b} $$

Verify with Graphs

To verify the result, we can graph the function and check that the limit as x approaches 0 from the positive side is indeed \(\frac{a}{b}\) for specific values of parameters a and b. For example, you can set a=2 and b=1 or a=3 and b=2, and plot the function, $$\left(a^{x}-b^{x}\right)^{1 / x}$$ If the graph approaches \(\frac{a}{b}\) when x approaches 0 from the positive side, our solution is correct.

Key concepts

L'Hopital's Rule

L'Hopital's Rule is a powerful tool for evaluating limits that result in indeterminate forms, such as 0/0 or ∞/∞. When you encounter a limit that gives an indeterminate form, you can apply this rule by differentiating the numerator and denominator separately and then taking the limit again. If the new limit is determinate, then it is the value of the original limit. Remember, the conditions for applying L'Hopital's Rule are: both the numerator and denominator must be differentiable, and the limit of the denominator must not be zero after differentiating.

For instance, in our exercise, we applied L'Hopital's Rule to \( \lim _{x \rightarrow 0^{+}}\frac{\ln(a^x)-\ln(b^x)}{x} \) which was an indeterminate form of type 0/0. After differentiating both the numerator and the denominator with respect to x, we were able to find a determinate limit and thus solve the problem.

Indeterminate Forms

Indeterminate forms occur in calculus when evaluating a limit leads to a result that isn't immediately apparent, such as 0/0, ∞/∞, 0∙∞, ∞ - ∞, 1^∞, 0^0, or ∞^0. These expressions do not have a well-defined value and require further mathematical tools, like L'Hopital's Rule or algebraic manipulation, to resolve. The exercise provided deals specifically with the indeterminate form of 0/0. It's important to recognize these forms early on because they indicate the need for additional steps to evaluate the limit, such as simplifying the expression or using a rule like L'Hopital's to find the correct value of the limit.

To improve understanding, recognizing the structure of indeterminate forms when faced with a limit problem is crucial. This identification often directs the route one must take for solving the limit.

Properties of Logarithms

Logarithms are incredibly versatile and have properties that make them useful for simplifying complex expressions, especially in limit problems. Some key properties include:

• The logarithm of a product is the sum of the logarithms: \(\ln(ab) = \ln a + \ln b\).
• The logarithm of a quotient is the difference of the logarithms: \(\ln(\frac{a}{b}) = \ln a - \ln b\).
• The logarithm of a power is the exponent times the logarithm: \(\ln(a^x) = x\ln a\).

These properties were crucial in the step-by-step solution as they allowed us to rewrite the exponential function in a form that made it easier to apply L'Hopital's Rule.

Understanding and applying logarithm properties can dramatically simplify the process of limit evaluation, particularly when dealing with exponential functions.

Exponential Functions

Exponential functions, of the form \( f(x) = a^x \), are fundamental in mathematics due to their unique properties and applications across various fields. They are easily recognizable by their constant base raised to a variable exponent. In calculus, these functions are smoothly continuous and differentiable, which implies that their slope varies at every point, and they have a well-defined rate of change at any point.

During the limit evaluation process, it's crucial to remember that \( e^{\ln a} = a \), based on the inverse properties of logarithms and exponentials. This relationship simplifies expressions and solves limits involving exponential functions, as seen in our exercise's final steps. Understanding the behavior and properties of exponential functions is essential to solving limits and will also be beneficial when exploring more advanced topics like differential equations and complex number theory.