Problem 42 - Calculus Early Transcendentals

Question

Evaluate the following limits. $$\lim _{x \rightarrow \infty} \frac{\ln \left(3 x+5 e^{x}\right)}{\ln \left(7 x+3 e^{2 x}\right)}$$

Step-by-Step Solution

Answer

$$\lim _{x \rightarrow \infty} \frac{\ln \left(3 x+5 e^{x}\right)}{\ln \left(7 x+3 e^{2 x}\right)}$$ Answer: The limit of the given expression as x approaches infinity is 1.

Step 1: Apply L'Hôpital's Rule

Since the given limit \(\lim _{x \rightarrow \infty} \frac{\ln \left(3 x+5 e^{x}\right)}{\ln \left(7 x+3 e^{2 x}\right)}\) is an indeterminate form of type \(\frac{\infty}{\infty}\), we can apply L'Hôpital's rule. Take the derivative of both numerator and denominator with respect to x: $$\frac{d}{dx}\left(\ln \left(3x + 5e^x\right)\right) = \frac{3 + 5e^x}{3x + 5e^x}$$ and $$\frac{d}{dx}\left(\ln \left(7x + 3e^{2x}\right)\right) = \frac{7 + 6e^{2x}}{7x + 3e^{2x}}$$ Applying L'Hôpital's rule, we have: $$\lim _{x \rightarrow \infty} \frac{3 + 5e^x}{3x + 5e^x} \cdot \frac{7x + 3e^{2x}}{7 + 6e^{2x}}$$

Step 2: Simplify the expression

Observe that we have the term \(5e^x\) in both the top and bottom and we can divide both numerator and denominator by \(5e^x\): $$\lim _{x \rightarrow \infty} \frac{3/5e^x + 1}{3x/5e^x + 1} \cdot \frac{7x/3e^{2x} + 1}{7/6e^{2x} + 1}$$

Step 3: Evaluate the limit

Now we can evaluate the limit for each term separately: $$\lim_{x \rightarrow \infty} \frac{3/5e^x + 1}{3x/5e^x + 1} \cdot \frac{7x/3e^{2x} + 1}{7/6e^{2x} + 1} = \lim_{x \rightarrow \infty} \frac{3e^{-x} + 1}{3xe^{-x} + 1} \cdot \frac{7xe^{-2x} + 1}{7e^{-2x} + 1}$$ As \(x \rightarrow \infty\), each of the exponential terms with negative exponents will tend to 0. Therefore, we have: $$= \frac{0 + 1}{0 + 1} \cdot \frac{0 + 1}{0 + 1} = 1 $$ Hence, the limit of the given expression is 1: $$\lim _{x \rightarrow \infty} \frac{\ln \left(3 x+5 e^{x}\right)}{\ln \left(7 x+3 e^{2 x}\right)} = 1$$

Key Concepts

Limits at Infinity

When we talk about limits at infinity, we're looking at what value a function approaches as the input becomes extremely large or extremely negative. This concept is critical in understanding how functions behave when their input grows without bounds.

For example, consider the expression \( \lim_{x \rightarrow \infty} \frac{1}{x} \). As \( x \) becomes larger and larger, the value of \( \frac{1}{x} \) becomes closer to 0. Hence, we say the limit of \( \frac{1}{x} \) as \( x \) approaches infinity is 0. This behavior can help us determine the long-term behavior of functions and their asymptotic tendencies. In calculus, understanding limits at infinity is essential for analyzing the behavior of functions and finding horizontal asymptotes.

Indeterminate Forms

In calculus, indeterminate forms arise when evaluating a limit leads to an expression that is not immediately obvious in determining the limit's value. One common type of indeterminate form is \( \frac{\infty}{\infty} \), which we encounter when both the numerator and the denominator of a fraction tend towards infinity.

Indeterminate forms like \( 0\times\infty \) or \( \frac{0}{0} \) can also occur, and they do not have inherent values without further manipulation. To resolve indeterminate forms and find the actual limit, techniques such as L'Hôpital's Rule, algebraic simplification, or finding a common denominator may be used. Identifying and resolving indeterminate forms is an essential skill in calculus, as it allows us to deal with complex limits and understand the behavior of functions.

Derivative of Logarithmic Functions

The derivative of logarithmic functions is a fundamental concept in calculus, especially when dealing with limits involving logarithms. To find the derivative of a logarithmic function, such as \( \ln(x) \), we use the formula \( \frac{d}{dx}\ln(x) = \frac{1}{x} \).

For more complex logarithmic expressions like \( \ln(3x + 5e^{x}) \), we apply the chain rule. This involves finding the derivative of the outer function (the logarithm itself) and multiplying it by the derivative of the inner function (the argument of the logarithm). This calculation plays a crucial role in applying L'Hôpital's Rule because we often differentiate logarithmic functions in the numerator and denominator to resolve indeterminate forms like \( \frac{\infty}{\infty} \). Comprehending the derivative of logarithmic functions is imperative for successful calculus problem-solving involving limits, growth and decay models, and integrations.