Question

Begin by plotting the function in an appropriate window. Your computer may indicate that some of these limits do not exist, but, if so, you should be able to interpret the answer as either \(\infty\) or \(-\infty\). $$ \lim _{x \rightarrow 0^{+}}(1+\sqrt{x})^{1 / \sqrt{x}} $$

Short answer

The limit is \(e\).

Step-by-step solution

Understanding the Problem

We need to find the limit of the expression \((1+\sqrt{x})^{1/\sqrt{x}}\) as \(x\) approaches 0 from the positive side. This involves analyzing the behavior of the function near 0.

Rewrite the Expression

Consider the term \((1+\sqrt{x})^{1/\sqrt{x}}\). This can be rewritten using the exponential function: \(e^{\ln((1+\sqrt{x})^{1/\sqrt{x}})}\). Simplify the exponent: \(e^{\frac{1}{\sqrt{x}} \cdot \ln(1+\sqrt{x})}\).

Apply L'Hôpital's Rule

The expression inside the exponential becomes \(\frac{\ln(1+\sqrt{x})}{\sqrt{x}}\). As \(x \rightarrow 0^{+}\), both numerator and denominator approach 0, allowing us to apply L'Hôpital's Rule. Differentiate: numerator \(\ln(1+\sqrt{x})' = \frac{1}{1+\sqrt{x}} \cdot \frac{1}{2\sqrt{x}}\) and denominator \(\sqrt{x}' = \frac{1}{2\sqrt{x}}\).

Simplify the Derivatives

After applying L'Hôpital's Rule, the expression becomes \(\frac{1}{1+\sqrt{x}}\). As \(x \rightarrow 0^{+}\), this tends to \(\frac{1}{1+0} = 1\).

Find the Exponential Limit

Since the expression inside the exponential limits to 1, the entire original expression limits to \(e^1 = e\).

Conclusion

Therefore, the limit \(\lim _{x \rightarrow 0^{+}}(1+\sqrt{x})^{1/\sqrt{x}} = e\).

Key concepts

L'Hôpital's Rule

L'Hôpital's Rule is a powerful tool in calculus, used to evaluate limits of indeterminate forms like \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \). When you encounter a limit problem where both the numerator and the denominator approach 0 or infinity, L'Hôpital's Rule can be applied to simplify the result.

Here is how it works:

• If \( \lim_{x \to c} f(x) = 0 \) and \( \lim_{x \to c} g(x) = 0 \) or both are infinity, then \( \lim_{x \to c} \frac{f(x)}{g(x)} = \lim_{x \to c} \frac{f'(x)}{g'(x)} \), provided the limit on the right-hand side exists.

Using this rule requires differentiating the numerator and denominator separately. This is a method of simplifying complex fractions that do not initially seem to have a straightforward limit.

In the provided solution, L'Hôpital's Rule helped evaluate the limit of the expression \( \frac{\ln(1+\sqrt{x})}{\sqrt{x}} \) as \( x \to 0^+ \). After applying the rule, the resulting expression \( \frac{1}{1+\sqrt{x}} \) could easily be evaluated.

Exponential Function

The exponential function, denoted by \( e^x \), is a fundamental function in calculus known for its unique property of being its own derivative. This function is crucial when dealing with continuous growth or decay and appears frequently in limit problems.

The equation \( e^{\ln((1+\sqrt{x})^{1/\sqrt{x}})} \) in our step-by-step solution showcases the transformation of the original expression into a form more amenable to limit evaluation. This transformation relies on the identity \( a^b = e^{b \cdot \ln(a)} \), leveraging the properties of natural logarithms to simplify complex power expressions.

By rewriting \((1+\sqrt{x})^{1/\sqrt{x}}\) as an exponential function, we unlocked a pathway to apply both logarithmic differentiation and L'Hôpital's Rule effectively. This approach simplifies the expression inside the limit, eventually leading us to the elegant result of \( e \).

Derivatives

Derivatives are a core concept in calculus, determining the rate at which a function changes at any given point. Understanding derivatives is crucial for applying L'Hôpital's Rule, as it involves taking derivatives of both the numerator and the denominator.

In our problem:

• The derivative of \( \ln(1+\sqrt{x}) \) is found using the chain rule, yielding \( \frac{1}{1+\sqrt{x}} \cdot \frac{1}{2\sqrt{x}} \).
• The derivative of \( \sqrt{x} \) is \( \frac{1}{2\sqrt{x}} \).

By substituting these derivatives into the expression \( \frac{\ln(1+\sqrt{x})}{\sqrt{x}} \), L'Hôpital's Rule facilitated the simplification process. The resulting expression \( \frac{1}{1+\sqrt{x}} \) is much simpler to analyze as \( x \to 0^+ \), allowing us to easily determine that the limit approaches 1. Mastery of derivative calculations underpins many advanced calculus techniques.

Limits

Limits are a foundational part of calculus, used to describe the behavior of functions as they approach a certain point, either finitely or infinitely. Understanding limits helps in determining the continuity and behavior of functions at boundary points.

For the equation \( \lim_{x \to 0^+} (1+\sqrt{x})^{1/\sqrt{x}} \), we are investigating what this expression approaches as \( x \) gets closer to 0 from the positive side.

Several key aspects involve analyzing limits:

• The expression must often be rewritten in a form that avoids indeterminate cases.
• Mathematically robust methods like L'Hôpital's Rule or logarithmic transformations enable effective limit calculation.

Solving limits often reveals a unique insight into the function’s behavior near specific points, unveiling solutions like \( e \) for our original problem. Comprehension of limits is thus central to both elementary and advanced calculus concepts.

Mathematical Analysis

Mathematical analysis is the branch of mathematics concerned with limits, continuity, infinity, and functions' behaviors. It delves into the deeper properties of calculus, fostering a comprehensive understanding of complex mathematical concepts.

Lessons from our exercise involve several analytical skills:

• Transforming expressions using natural logarithms and exponential functions to simplify calculations.
• Applying L'Hôpital's Rule to navigate indeterminate forms.
• Employing derivative techniques to simplify and evaluate limits effectively.

Through mathematical analysis, one gains the ability to dissect intricate problems, revealing elegant solutions like that in our problem where the limit resulted in \( e \). Analysis equips learners with insights and tools to approach new challenges systematically. Understanding such principles underpins much of modern mathematics and its applications.