Problem 19 - Calculus for Biology and Medicine

Question

Evaluate the limits. $$ \lim _{x \rightarrow \infty} \frac{3 e^{2 x}+1}{2 e^{2 x}-e^{x}} $$

Step-by-Step Solution

Answer

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

Step 1: Identify Form of Limit

The limit expression \( \frac{3 e^{2x}+1}{2 e^{2x}-e^x} \) as \( x \to \infty \) involves exponential functions. As \( x \to \infty \), both the numerator and the denominator tend toward infinity, leading to an indeterminate form of type \( \frac{\infty}{\infty} \).

Step 2: Factor Out the Dominant Term

In expressions involving exponentials, the term with the highest power usually dominates. Here, \( e^{2x} \) is the dominant term in both the numerator and the denominator. Factor \( e^{2x} \) out from both the top and the bottom:\[ \frac{3e^{2x}+1}{2e^{2x}-e^x} = \frac{e^{2x}(3 + \frac{1}{e^{2x}})}{e^{2x}(2 - \frac{e^x}{e^{2x}})}. \]

Step 3: Simplify the Expression

After factoring, the \( e^{2x} \) terms in the numerator and denominator cancel out:\[ \frac{3 + \frac{1}{e^{2x}}}{2 - \frac{1}{e^x}}. \]

Step 4: Evaluate the Limit

As \( x \to \infty \), both \( \frac{1}{e^{2x}} \to 0 \) and \( \frac{1}{e^x} \to 0 \). Substitute these limits into the expression:\[ \lim_{x \to \infty} \frac{3 + \frac{1}{e^{2x}}}{2 - \frac{1}{e^x}} = \frac{3 + 0}{2 - 0} = \frac{3}{2}. \]

Key Concepts

Indeterminate Forms

When evaluating limits, we sometimes encounter expressions that at first glance seem undefined or difficult to evaluate. These are known as indeterminate forms.

• Common indeterminate forms include \( \frac{0}{0} \), \( \frac{\infty}{\infty} \), \( \infty - \infty \), and others.
• These forms occur because both the numerator and denominator (or parts of an expression) approach some extreme values that don't directly provide a limit.

In the given exercise, the expression is \( \frac{3 e^{2x} + 1}{2 e^{2x} - e^{x}} \). As \( x \to \infty \), both the numerator and the denominator go toward infinity.
This results in an indeterminate form of \( \frac{\infty}{\infty} \).
To resolve this, we need to simplify the expression so that the limit becomes apparent. Factoring out the dominant exponential terms is a common technique that helps eliminate indeterminate forms.

Exponential Functions

Exponential functions are critical in mathematics, especially when studying growth processes and calculating limits as variables approach extreme values.

• The function \( e^{x} \) is the standard exponential function which grows incredibly fast as \( x \to \infty \).
• In our expression, \( e^{2x} \) and \( e^{x} \) are the terms causing the numerator and denominator to tend towards infinity.

A powerful technique when dealing with exponential expressions is to factor out the dominant exponential term.
In this case, \( e^{2x} \) appears in both the numerator and the denominator. By factoring it out, we simplify the limit to another form that can be evaluated without indeterminate issues.
This technique is particularly useful in calculus for making expressions easier to handle and simplifying limits that at first appear complex.

Asymptotic Behavior

Understanding asymptotic behavior is vital when dealing with functions that tend towards infinite values. This behavior reveals what happens to a function as the input grows close to infinity or some finite value.

• In the context of limits, asymptotic behavior helps understand how a function behaves at its extremes, such as near a vertical or horizontal asymptote.
• The concept often connects with exponential functions, as these can show rapid growth, overshadowing less dominant terms as \( x \to \infty \).

In the problem, we look at the asymptotic behavior of the expression \( \frac{3 + \frac{1}{e^{2x}}}{2 - \frac{1}{e^x}} \) after simplification.
As \( x \to \infty \), the terms \( \frac{1}{e^{2x}} \to 0 \) and \( \frac{1}{e^{x}} \to 0 \), simplifying the behaviors of the numerator and denominator as constants: \( 3 \) and \( 2 \) respectively.
Thus, the asymptotic behavior ensures that the limit is simply \( \frac{3}{2} \), predicting how the function balances out as \( x \) becomes exceedingly large.