Problem 1 - A Primer of Real Analysis

Question

Use l'Hôpital's rule to compute $$ \lim _{x \rightarrow 0^{+}} \frac{\sqrt{1+x}-1}{x} $$

Step-by-Step Solution

Answer

The limit is \( \frac{1}{2} \).

Step 1: Identify form of limit

First, identify the form of the limit \( \lim_{x \to 0^+} \frac{\sqrt{1+x} - 1}{x} \). As \( x \to 0^+ \), both the numerator \( \sqrt{1+x} - 1 \to 0 \) and the denominator \( x \to 0 \). Therefore, this limit is of the indeterminate form \( \frac{0}{0} \), which allows us to use l'Hôpital's Rule.

Step 2: Differentiate the numerator and denominator

According to l'Hôpital's Rule, we differentiate the numerator and the denominator of \( \frac{\sqrt{1+x} - 1}{x} \).The derivative of the numerator \( \sqrt{1+x} - 1 \) with respect to \( x \) is \( \frac{1}{2\sqrt{1+x}} \).The derivative of the denominator \( x \) with respect to \( x \) is \( 1 \).

Step 3: Apply l'Hôpital's Rule

Now, apply l'Hôpital's Rule to the original limit:\[ \lim_{x \to 0^+} \frac{\sqrt{1+x} - 1}{x} = \lim_{x \to 0^+} \frac{\frac{1}{2\sqrt{1+x}}}{1} \]This simplifies to:\[ \lim_{x \to 0^+} \frac{1}{2\sqrt{1+x}} \]

Step 4: Evaluate the simplified limit

Evaluate the limit \( \lim_{x \to 0^+} \frac{1}{2\sqrt{1+x}} \) as \( x \to 0^+ \). When \( x \to 0 \), \( \sqrt{1+x} \to \sqrt{1} = 1 \). Therefore, the expression becomes \( \frac{1}{2(1)} = \frac{1}{2} \).

Key Concepts

Indeterminate Forms

In calculus, the concept of indeterminate forms is crucial for understanding limits that aren't straightforward to evaluate. An indeterminate form arises when an expression takes on an ambiguous form, particularly as it approaches certain points. For example, consider the form \( \frac{0}{0} \). It isn’t clearly defined mathematically because both the numerator and denominator approach zero, leading to uncertainty.

• The form \( \frac{0}{0} \) often appears in calculus problems, especially when dealing with limits of functions at specific points.
• Indeterminate forms signal the need for special techniques, like l'Hôpital's Rule, to evaluate the actual limit.

Recognizing indeterminate forms helps determine when to apply methods like differentiation to properly analyze and solve limit problems. By systematically identifying these forms, you can use appropriate mathematical rules to find a definite value for the limit.

Derivatives

Derivatives, a fundamental concept in calculus, represent the rate of change of a function with respect to a variable. When we talk about the derivatives in the context of l'Hôpital's Rule, we're leveraging their ability to analyze small changes in functions.

• The derivative of a function provides a new function that describes how the original function changes as its input changes.
• In the context of limits, derivatives help us re-evaluate expressions that have become indeterminate like \( \frac{0}{0} \).

In solving \( \lim_{x \to 0^+} \frac{\sqrt{1+x} - 1}{x} \), the derivative of the numerator \( \sqrt{1+x} - 1 \) is \( \frac{1}{2\sqrt{1+x}} \) and the denominator \( x \) is simply \( 1 \).
By replacing the original limit expression with the derivatives of the top and bottom, the problem becomes much simpler to solve, overcoming the indeterminate form.

Limits in Calculus

Limits are a central idea in calculus, used to define derivatives and integrals. They help in understanding the behavior of functions as they approach specific points, particularly where direct substitution is not possible or leads to ambiguity.

• The limit of a function describes the value it approaches as the input approaches a certain point.
• Calculating limits where the direct substitution leads to indeterminate forms often requires strategies like simplification or using l'Hôpital's Rule.

In the problem \( \lim_{x \to 0^+} \frac{\sqrt{1+x} - 1}{x} \), the direct substitution of \( x = 0 \) into both the numerator and the denominator results in the indeterminate form \( \frac{0}{0} \). This is where the limit can be reevaluated using derivatives.
Once simplified using derivatives, the new limit \( \lim_{x \to 0^+} \frac{1}{2\sqrt{1+x}} \) evaluates to \( \frac{1}{2} \) as \( x \to 0 \), demonstrating how limits reveal the behavior of a function at specific points.