Question

Evaluate the limits with either L'Hôpital's rule or previously learned methods. $$ \lim _{x \rightarrow 0} \frac{3^{x}-2^{x}}{x} $$

Short answer

The limit is \(\ln\left(\frac{3}{2}\right)\).

Step-by-step solution

Check Indeterminate Form

First, we need to confirm if the limit has an indeterminate form that allows us to use L'Hôpital's Rule. When we substitute \(x = 0\) in the expression \(\frac{3^x - 2^x}{x}\), both the numerator \(3^x - 2^x\) and the denominator \(x\) approach zero. Thus, we have the \(\frac{0}{0}\) indeterminate form, and we can apply L'Hôpital's Rule.

Apply L'Hôpital's Rule

L'Hôpital's Rule states that if the limit \(\lim _{x \rightarrow c} \frac{f(x)}{g(x)}\) results in an indeterminate form \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\), then the limit can be found using \(\lim _{x \rightarrow c} \frac{f'(x)}{g'(x)}\), if this limit exists. Here, the function in the numerator is \(f(x) = 3^x - 2^x\) and the denominator is \(g(x) = x\). We differentiate each: \(f'(x) = 3^x \, \ln(3) - 2^x \, \ln(2)\) and \(g'(x) = 1\).

Evaluate New Limits

Now, re-evaluate the limit: \(\lim _{x \rightarrow 0} \frac{3^x \, \ln(3) - 2^x \, \ln(2)}{1}\). Since both \(3^x\) and \(2^x\) approach 1 as \(x\) approaches 0, we compute \(3^0 = 1\) and \(2^0 = 1\). Therefore, the expression simplifies to \(\ln(3) - \ln(2)\).

Compute the Final Result

Calculate the numerical value of the expression: \(\ln(3) - \ln(2)\). This can be represented as \(\ln\left(\frac{3}{2}\right)\). Hence, the limit is \(\ln\left(\frac{3}{2}\right)\).

Key concepts

Indeterminate Forms

In calculus, understanding indeterminate forms is essential as it indicates cases where the computation of limits involves expressions like \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \), which do not directly provide a specific limit value. These forms are often encountered in the evaluation of limits and include other variations like \( 0 \times \infty \), \( \infty - \infty \), and \( 1^{\infty} \). When faced with these forms, simple substitution doesn't suffice to solve the limit problem. To resolve indeterminate forms, you can apply techniques like:

• Factoring and simplifying expressions.
• Using L'Hôpital's Rule for forms \( \frac{0}{0} \) and \( \frac{\infty}{\infty} \).
• Using algebraic manipulation or trigonometric identities.
• Changing the form of the function using continuous functions.

These techniques are used to transform the expression into a more manageable form and to calculate the limit effectively. L'Hôpital's Rule is particularly powerful for cases of \( \frac{0}{0} \) and \( \frac{\infty}{\infty} \) since it allows differentiating the numerator and denominator separately to find the new limit.

Exponential Functions

Exponential functions, like \( a^x \) where \( a \) is a positive real number, play a crucial role in calculus due to their unique properties. They involve growth or decay processes that depend on the power of a constant base \( a \). Some important characteristics of exponential functions include:

• Continuous and differentiable for all real numbers.
• Have a constant rate of relative growth proportional to the function itself.
• The derivative of \( a^x \) concerning \( x \) is \( a^x \ln(a) \).

Exponential functions often appear in limit problems because they can rapidly increase or decrease, creating situations where direct substitution leads to indeterminate forms. In these cases, understanding their derivatives can be crucial, especially when using L'Hôpital's Rule to find limits. In the original exercise, both \( 3^x \) and \( 2^x \) are exponential functions whose properties allow them to be differentiated into usable forms for solving limits.

Limits in Calculus

Limits in calculus are fundamental as they describe the behavior of functions as they approach a certain point. Evaluating limits is crucial for understanding concepts such as continuity, derivatives, and integrals. Calculating limits involves determining what value a function approaches as the input gets arbitrarily close to a particular point. There are several ways to evaluate limits, including:

• Direct substitution, whenever possible, if no indeterminate form arises.
• Factoring and canceling.
• Simplifying complex expressions.
• Using special techniques like L'Hôpital's Rule.

L'Hôpital's Rule specifically helps when direct substitution leads to \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \). It involves differentiating the numerator and the denominator separately to find a new limit that can often be directly computed. Limits are crucial not only in theoretical mathematics but also in practical applications like physics, engineering, and economics, where they model various changes and rates over time.