Problem 16 - Calculus for Biology and Medicine

Question

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

Step-by-Step Solution

Answer

The limit evaluates to 1.

Step 1: Identify the Indeterminate Form

Begin by looking at the expression \[ \lim _{x \rightarrow \infty} \frac{1-e^{x}}{2-e^{x}}. \] As \(x\) approaches infinity, \(e^x\) tends to infinity, making both the numerator and the denominator have the form \( -\infty \). Thus, the expression is an indeterminate form \(\frac{-\infty}{-\infty}\).

Step 2: Simplify the Expression

Divide both the numerator and the denominator by \(e^x\) to simplify:\[ \lim _{x \rightarrow \infty} \frac{1/e^{x} - 1}{2/e^{x} - 1}. \]Now evaluate each term as \(x\) approaches infinity.

Step 3: Evaluate Each Term as x Approaches Infinity

Notice that as \(x\) approaches infinity:- \(\frac{1}{e^x} \rightarrow 0\).Thus, the expression simplifies to \[ \lim _{x \rightarrow \infty} \frac{0 - 1}{0 - 1} = \frac{-1}{-1}. \]

Step 4: Calculate the Limit

The simplified limit is \[ \frac{-1}{-1} = 1. \]

Step 5: Conclusion

Hence, \[ \lim _{x \rightarrow \infty} \frac{1-e^{x}}{2-e^{x}} = 1. \]

Key Concepts

Indeterminate Forms

In calculus, an indeterminate form occurs when evaluating a limit yields a result that isn't instantly clear, such as \( \frac{-\infty}{-\infty} \). This can be confusing because both the numerator and the denominator grow to infinity at the same pace, suggesting the expression might resolve to any number. Recognizing indeterminate forms is crucial because they signal that more steps are needed to find the true limit.
Common indeterminate forms include:

• \( \frac{0}{0} \)
• \( \frac{\pm \infty}{\pm \infty} \)
• \( 0 \cdot \infty \)
• \( \infty - \infty \)
• \( 1^{\infty}, \infty^0, 0^0 \)

Identifying these forms allows you to apply techniques to resolve them into a definitive answer. For the given exercise, the form \( \frac{-\infty}{-\infty} \) required simplification to reach the solution.

Exponential Functions

Exponential functions, such as \( e^x \), are vital in calculus because they grow rapidly and have unique properties. The base \( e \) approximately equals 2.718, known as Euler's number. An important property of \( e^x \) is that its rate of growth increases as \( x \) becomes larger.
This makes \( e^x \) particularly noteworthy when evaluating limits as \( x \) approaches infinity, as seen in the original exercise. Understanding these functions helps in predicting behavior of expressions as they tend towards infinity or zero.

• \( e^x \) grows without bound as \( x \) approaches infinity, which affects both the numerator and the denominator in fractional expressions.
• Additionally, as \( x \) approaches negative infinity, \( e^x \) approaches zero.

These behaviors make \( e^x \) very predictable, which aids in limit evaluation.

Limit Evaluation Techniques

Evaluating limits is a fundamental concept in calculus used to predict the behavior of functions as they approach certain points. For indeterminate forms, one of the most common limit evaluation techniques is simplifying the expression.
For the exercise provided, the technique used was dividing both the numerator and the denominator by \( e^x \), which simplifies the expression to a form that can be directly evaluated.
Another effective technique is L'Hôpital's Rule, which applies to forms like \( \frac{0}{0} \) and \( \frac{\infty}{\infty} \). This rule suggests taking the derivative of the numerator and the denominator to simplify the evaluation.

• Always search for algebraic manipulation first to avoid unnecessary complexity.
• Prioritize simplifying expressions by factoring out leading terms or canceling common factors.

Mastering these techniques ensures you can tackle a wide variety of limit problems with confidence.