Problem 49 - Single Variable Calculus

Question

Evaluate $$ \lim _{n \rightarrow \infty} \frac{1-\left(1-\frac{1}{n}\right)^{9}}{1-\left(1-\frac{1}{n}\right)} $$ Hint: Use Theorem 1 .

Step-by-Step Solution

Answer

The limit of the given function as \(n\) approaches infinity is \(-9\), found by applying L'Hopital's Rule to the indeterminate form \(\frac{0}{0}\) and finding the limit of the ratio of the derivatives of the numerator and denominator functions.

Step 1: Verify it is an Indeterminate Form

To use L'Hopital's Rule, first verify that the given function has an indeterminate form as \(n\) approaches infinity. Plugging in \(n = \infty\) in the given function, we get: \[ \lim _{n \rightarrow \infty}\frac{1-\left(1-\frac{1}{n}\right)^{9}}{1-\left(1-\frac{1}{n}\right)} = \frac{1-(1-0)^{9}}{1 - (1-0)} = \frac{0}{0} \] Since we have the indeterminate form \(\frac{0}{0}\), we can proceed with L'Hopital's Rule to solve the problem.

Step 2: Define the Numerator and the Denominator Functions of the Limit

Let's define the numerator function, \(N(n)\),and the denominator function, \(D(n)\), as: \(N(n) = 1 - (1 - \frac{1}{n})^9\) \(D(n) = 1 - (1 - \frac{1}{n})\)

Step 3: Derive the First Derivatives of N(n) and D(n)

Now, we will find the derivatives of these two functions with respect to \(n\). \(N'(n) = \frac{d}{dn}\left[1 - (1 - \frac{1}{n})^9\right] = -9(1 - \frac{1}{n})^8 \cdot \frac{-1}{n^2}\) \(D'(n) = \frac{d}{dn}\left[1 - (1 - \frac{1}{n})\right] = \frac{1}{n^2}\)

Step 4: Apply L'Hopital's Rule and Find the Limit

Now, we apply L'Hopital's Rule and find the limit of the ratio of the derivatives: \[ \lim _{n \rightarrow \infty} \frac{N'(n)}{D'(n)} = \lim _{n \rightarrow \infty} \frac{-9(1 - \frac{1}{n})^8 \cdot \frac{-1}{n^2}}{\frac{1}{n^2}} \] Since both derivatives have a common factor, which is \(\frac{1}{n^2}\), we can simplify the expression: \[ \lim _{n \rightarrow \infty} -9(1 - \frac{1}{n})^8 = -9(1 - 0)^8 = -9 \] So, the limit of the original expression is \(-9\).

Key Concepts

Indeterminate Form

In calculus, an *indeterminate form* surfaces when evaluating a limit results in an expression that does not have a straightforward, defined value. One common type is the \( \frac{0}{0} \) form. This occurs when both the numerator and the denominator approach zero. It's important because it suggests that direct substitution might not work, and alternative methods, like L'Hopital's Rule, might be necessary to find the limit.

To identify this in the problem, we substituted \( n = \infty \) into the functions. This gave the indeterminate form \( \frac{0}{0} \), showcasing the need for a specific strategy to solve the limit problem. Indeterminate forms are crucial in calculus as they signal ambiguity in limit values, prompting us to dig deeper into the behavior around the limit point.

Calculus Limit

A *calculus limit* helps us understand the behavior of a function as it approaches a particular point. In this exercise, we are interested in what happens as \( n \rightarrow \infty \). Limits are fundamental in calculus because they allow us to cope with infinity and zero in precise ways.

Calculating the limit of a function may not always be direct. Some expressions simplify when plugged directly into the limit, revealing clear values. Others, like our indeterminate form, require techniques like L'Hopital's Rule to make sense of the expression. Using limits, we can define derivatives and integrals, making them the backbone concepts of calculus.

• The limit gives insight into how functions behave at boundaries, such as infinity or specific points.
• Limits are foundational for defining derivatives, which are about understanding instantaneous change.

Differentiation

*Differentiation* is the process of finding the derivative of a function, which tells us the rate at which one quantity changes with respect to another. In this exercise, differentiation played a big role when using L'Hopital's Rule. This rule states that if a fraction is in the indeterminate form \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \), the limit of the fraction can be found by differentiating the numerator and the denominator separately and then taking the limit again.

For our exercise, we derived the functions \( N(n) \) and \( D(n) \). After finding their derivatives, we simplified the expression and successfully determined the limit resulted in \( -9 \).

• Differentiation allows us to understand how functions change at any point.
• In L'Hopital's Rule, it transforms an indeterminate form into a computable limit.
• Understanding differentiation lets us tackle complex calculus problems effectively.

Differentiation is key in calculus as it breaks down dynamic changes, underpinning much of the subject's applications.