Question

Evaluate the limits with either L'Hôpital's rule or previously learned methods. $$ \lim _{x \rightarrow 0} \frac{(1+x)^{n}-1}{x} $$

Short answer

The limit is \( n \).

Step-by-step solution

Identify the Form

First, examine the limit expression to determine if it's an indeterminate form. Here, as \( x \to 0 \), the expression \( (1+x)^n - 1 \) approaches \( n \cdot 0 = 0 \) and the denominator \( x \) also approaches 0. Hence, it takes an indeterminate form of \( \frac{0}{0} \).

Apply L'Hôpital's Rule

Since we have an indeterminate form \( \frac{0}{0} \), we can apply L'Hôpital's Rule. This rule states to differentiate the numerator and the denominator separately and then take the limit again:\[\lim_{x \to 0} \frac{(1+x)^n - 1}{x} = \lim_{x \to 0} \frac{\frac{d}{dx}((1+x)^n - 1)}{\frac{d}{dx}(x)}\]Differentiating the numerator using the chain rule gives \( n(1+x)^{n-1} \), and differentiating the denominator gives 1.

Evaluate the New Limit

Substitute the derivatives into the limit:\[\lim_{x \to 0} \frac{n(1+x)^{n-1}}{1} = n \cdot \lim_{x \to 0} (1+x)^{n-1}\]As \( x \to 0 \), \( (1+x)^{n-1} \to (1)^{n-1} = 1 \). So, the limit evaluates to \( n \cdot 1 = n \).

Key concepts

Indeterminate Form

When evaluating limits, sometimes you encounter an expression that doesn't seem to work with simple algebraic manipulations. One common scenario is the indeterminate form. An indeterminate form doesn't have a directly computable value and often appears as forms like \( \frac{0}{0} \), \( \frac{\infty}{\infty} \), or \( 0 \cdot \infty \). In our exercise, as \( x \to 0 \), both the numerator \((1+x)^n - 1\) and the denominator \(x\) go to zero, resulting in an indeterminate form \(\frac{0}{0}\). This signals that we need a more sophisticated approach like L'Hôpital's Rule to evaluate the limit correctly.
Identifying an indeterminate form is the first critical step in determining when and how to apply calculus rules to find the limit of a function. Understanding this concept helps clear confusion where the limits can’t be resolved by simple substitution or algebraic simplifications.

Differentiation

Differentiation is the process of finding the derivative of a function. Derivatives indicate how the function value changes as its input changes. In the context of L'Hôpital's Rule, we differentiate the numerator and the denominator separately when faced with an indeterminate form.

In our example, differentiating the numerator \((1+x)^n - 1\) involves using the chain rule, which is essential when dealing with composite functions. It results in \( n(1+x)^{n-1} \). The denominator \(x\) differentiates to 1, which simplifies the limit evaluation.

• The derivative of \((1+x)^n\) is \( n(1+x)^{n-1} \)
• The derivative of the constant \(-1\) is 0
• The derivative of \(x\) is 1

This simplification step is crucial as it transforms the initial indeterminate form into an expression where the limit can be easily evaluated by substitution.

Limit Evaluation

Once the expression is reduced via differentiation, limit evaluation becomes straightforward. In essence, limit evaluation is the process of finding the value that a function approaches as the input approaches a particular point, often critical when analyzing the behavior of functions near undefined points.

After applying L'Hôpital's Rule, the limit simplifies to \( n \cdot \lim_{x \to 0} (1+x)^{n-1} \). When \( x \to 0 \), \( (1+x)^{n-1} \to 1 \), since any base raised to power zero equals 1. Therefore, we find that the limit evaluates to \( n \).
This step highlights the powerful utility of L'Hôpital's Rule in resolving complicated limits and provides a deeper understanding of the function's behavior as \( x \to 0 \). Mastery of these techniques allows for efficient problem-solving in calculus, especially for those indeterminate forms that standard algebra cannot handle easily.