Question

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

Short answer

The limit is \( e \).

Step-by-step solution

Understand the Expression as a Limit

The given limit is \( \lim_{x \to 0} (x+1)^{1/x} \). To solve it, observe the indeterminate form. As \( x \to 0 \), \(1/x \to \infty\) which can be challenging to evaluate directly.

Transform the Limit Expression

To simplify, consider the natural logarithm of the expression:\[ y = (x+1)^{1/x} \implies \ln(y) = \frac{1}{x} \ln(x+1) \].Now, solve \( \lim_{x \to 0} \ln(y) = \lim_{x \to 0} \frac{\ln(x+1)}{x} \).

Prepare to Apply L'Hôpital's Rule

The limit \( \lim_{x \to 0} \frac{\ln(x+1)}{x} \) is in the indeterminate form \( \frac{0}{0} \), which is suitable for L'Hôpital's Rule.

Apply L'Hôpital's Rule

Differentiate the numerator and the denominator:\( \frac{d}{dx} \ln(x+1) = \frac{1}{x+1} \) and \( \frac{d}{dx} x = 1 \).So the new limit becomes:\[ \lim_{x \to 0} \frac{1/(x+1)}{1} = \lim_{x \to 0} \frac{1}{x+1} \].

Evaluate the Limit

Substitute \( x = 0 \) into the limit:\[ \lim_{x \to 0} \frac{1}{x+1} = \frac{1}{1} = 1 \].Thus, \( \lim_{x \to 0} \ln(y) = 1 \).

Return to the Original Expression

Since \( \ln(y) = 1 \), then \( y = e^1 = e \).Therefore, \( \lim_{x \to 0} (x+1)^{1/x} = e \).

Key concepts

Limits

In calculus, limits are a fundamental concept that describe the behavior of a function as its input approaches a particular value. When you hear the phrase "the limit of a function as x approaches 0," it signifies understanding the value that the function nears, even though it may not physically "reach" that value.
The expression we are dealing with, \( \lim_{x \to 0}(x+1)^{1/x} \), truly benefits from the concept of limits.

• Approaching Value: Imagine \( x \) is getting closer and closer to 0. We want to know what happens to \((x+1)^{1/x}\) as this happens.
• Behavior Near Zero: Although \( (x+1)^{1/x} \) does not straightforwardly evaluate when \( x = 0 \), limits allow us to analyze the process at hand without direct substitution.

When evaluated using limits, the expression becomes more manageable with mathematical techniques like L'Hôpital's Rule or transformations like the natural logarithm.

Natural Logarithm

The natural logarithm, denoted by \( \ln(x) \), is a powerful tool in calculus beneficial for dealing with exponential growth, rates of change, and simplification of difficult expressions. When faced with an expression in the exponent, finding the natural logarithm can often transform it into a more workable form.

• Logarithmic Simplification: In our problem, transforming \((x+1)^{1/x}\) into \(\ln(y) = \frac{1}{x} \ln(x+1)\) makes the complex power expression more simplistic, as logarithms help unravel the exponent.
• Ease of Differentiation: After simplifying with the natural logarithm, it becomes much easier to apply differentiation -- a crucial step when using L'Hôpital's Rule.

When applying L'Hôpital's, the natural logarithm of the complex base is differentiated, leading to a simpler limit evaluation. This method reveals insights that direct evaluation may conceal.

Indeterminate Form

The idea of indeterminate forms is central to solving certain complicated limits. An indeterminate form, such as \( \frac{0}{0} \), indicates a mathematical "stalemate," where direct substitution does not result in a clear value. This is particularly common with limits involving division of two functions where both approach zero.

• Recognizing Indeterminacy: In our example, \( \lim_{x \to 0} \frac{\ln(x+1)}{x} \) forms a \( \frac{0}{0} \) scenario. This happens because \( \ln(x+1) \to \ln(1) = 0 \) and \( x \to 0 \), leading to an apparent division of zero by zero.
• Resolving with L'Hôpital's Rule: L'Hôpital's Rule, specifically designed to tackle indeterminate forms \( \frac{0}{0} \) and \( \frac{\infty}{\infty} \), allows us to differentiate the numerator and the denominator separately to find the viable limit.

Understanding and recognizing when an expression is in an indeterminate form prompts the application of techniques that turn a mathematical puzzle into a solvable problem.