Question

Find the indicated limit. Make sure that you have an indeterminate form before you apply l'Hôpital's Rule. $$ \lim _{x \rightarrow 0^{+}} \frac{7^{\sqrt{x}}-1}{2^{\sqrt{x}}-1} $$

Short answer

The limit is \( \frac{\ln(7)}{\ln(2)} \).

Step-by-step solution

Evaluate the Limit Directly

First, substitute the value of \( x \) approaching \( 0^{+} \) into the expression \( \frac{7^{\sqrt{x}}-1}{2^{\sqrt{x}}-1} \). This gives \( \frac{7^{0}-1}{2^{0}-1} = \frac{1-1}{1-1} = \frac{0}{0} \). The expression is in the indeterminate form \( \frac{0}{0} \), so we can apply l'Hôpital's Rule.

Apply l'Hôpital's Rule

Since the limit is in the form \( \frac{0}{0} \), apply l'Hôpital's Rule, which requires taking the derivatives of the numerator and the denominator. Remember the derivatives for exponential functions in the form \( a^{f(x)} \): if \( f(x) = \sqrt{x} \), then the derivative is \( a^{f(x)} \cdot \ln(a) \cdot \frac{d}{dx}[f(x)] \).

Derivative of the Numerator

Find the derivative of \( g(x) = 7^{\sqrt{x}} - 1 \). The derivative will be \( 7^{\sqrt{x}} \cdot \ln(7) \cdot \frac{1}{2\sqrt{x}} \).

Derivative of the Denominator

Find the derivative of \( h(x) = 2^{\sqrt{x}} - 1 \). The derivative will be \( 2^{\sqrt{x}} \cdot \ln(2) \cdot \frac{1}{2\sqrt{x}} \).

Simplify the Expression

Substitute the derivatives into the quotient: \[\lim _{x \rightarrow 0^{+}} \frac{7^{\sqrt{x}} \cdot \ln(7) \cdot \frac{1}{2\sqrt{x}}}{2^{\sqrt{x}} \cdot \ln(2) \cdot \frac{1}{2\sqrt{x}}}\] Cancel out \( \frac{1}{2\sqrt{x}} \) from the numerator and denominator as it is a common factor. We are left with \[\lim _{x \rightarrow 0^{+}} \frac{7^{\sqrt{x}} \cdot \ln(7)}{2^{\sqrt{x}} \cdot \ln(2)}.\] As \( x \rightarrow 0^{+} \), \( \sqrt{x} \rightarrow 0 \), making \( 7^{\sqrt{x}} \rightarrow 1 \) and \( 2^{\sqrt{x}} \rightarrow 1 \). Thus, the limit becomes \( \frac{\ln(7)}{\ln(2)} \).

Conclude the Limit

Evaluate the limit using the simplified expression: \[\lim _{x \rightarrow 0^{+}} \frac{7^{\sqrt{x}}-1}{2^{\sqrt{x}}-1} = \frac{\ln(7)}{\ln(2)}.\] This result is the indicated limit for the problem.

Key concepts

Indeterminate Forms

When evaluating limits, you might encounter what are called indeterminate forms. These are expressions that don’t initially resolve to a specific number and include forms like \( \frac{0}{0} \), \( \frac{\infty}{\infty} \), \( 0 \times \infty \), among others. An expression such as \( \frac{0}{0} \) doesn’t define a precise numerical value, creating a need for additional methods to solve them, such as l'Hôpital's Rule.

To identify an indeterminate form, substitute the variable's limiting value into the function. If substitution leads to something uncertain like \( \frac{0}{0} \), you’ve encountered such a form. This scenario was exactly what we found in our expression \( \lim _{x \rightarrow 0^{+}} \frac{7^{\sqrt{x}}-1}{2^{\sqrt{x}}-1} \), which simplified to \( \frac{0}{0} \).

Recognizing indeterminate forms is crucial as it tells us that the limit cannot be evaluated directly by substitution, and therefore, methods like l'Hôpital’s Rule, which involves differentiation, must be employed.

Exponential Functions

Exponential functions are mathematical expressions in which a constant base is raised to a variable exponent. These are typically written in the form \( a^x \), where \( a \) is the base and \( x \) is the exponent. They appear frequently in both pure and applied mathematics because of their unique properties, such as their continuous and smooth curve.

In our given limit problem, we saw exponential functions like \( 7^{\sqrt{x}} \) and \( 2^{\sqrt{x}} \). Here, the base numbers 7 and 2 are raised to the power of \( \sqrt{x} \), demonstrating how exponents can be functions themselves, like \( f(x) = \sqrt{x} \).

The derivative of such functions is crucial when applying l'Hôpital’s Rule. The general rule for finding the derivative of an exponential function with a function as an exponent, \( a^{f(x)} \), is to multiply the exponential function itself by the natural logarithm \( \ln(a) \) and the derivative of the exponent \( f(x) \). This technique was pivotal in our problem, allowing us to differentiate both the numerator and the denominator for limit evaluation.

Limit Evaluation

Limit evaluation is a fundamental concept in calculus, used to determine the value that a function approaches as the input approaches some value. When direct substitution in a limit gives an indeterminate form, special techniques such as l'Hôpital’s Rule become necessary.

The purpose of finding a limit is to predict the behavior of functions at points that might be undefined or difficult to analyze directly. In our exercise, we evaluated \( \lim _{x \rightarrow 0^{+}} \frac{7^{\sqrt{x}}-1}{2^{\sqrt{x}}-1} \), initially reducing to \( \frac{0}{0} \).

Applying l’Hôpital’s Rule involved taking the derivatives of both the numerator and denominator and simplifying the resulting expression. After simplification, the expressions \( 7^{\sqrt{x}} \, \rightarrow \, 1 \) and \( 2^{\sqrt{x}} \, \rightarrow \, 1 \) led us to compute the final limit. Understanding this process helps in seeing that limits help define otherwise ambiguous scenarios, providing clarity and concrete numerical outcomes such as \( \frac{\ln(7)}{\ln(2)} \).

Mastering these concepts allows you to tackle indeterminate forms and comprehend the behavior of complex functions near particular input values.