Evaluating Limit Of (1+x)^(1/n) - 1 / X As X Approaches 0

Introduction

In calculus, evaluating limits is a fundamental concept, and it's essential to understand how functions behave as their input approaches a specific value. In this article, we will delve into a particular limit problem that involves a radical expression. Specifically, we aim to evaluate the limit of the expression (1+x)^(1/n) - 1 / x as x approaches 0. This limit is a classic example that appears in various contexts, including the definition of derivatives and the study of power functions. We will explore different methods to solve this limit, including algebraic manipulation and L'Hôpital's Rule, providing a comprehensive understanding of the solution. The ability to evaluate such limits is crucial for students and professionals in mathematics, physics, engineering, and other related fields. Understanding how to deal with these types of expressions is also vital for tackling more complex problems in calculus and analysis. This discussion will not only present the solution but also provide a detailed explanation of the underlying principles and techniques involved, ensuring a solid grasp of the material.

Understanding the Limit Problem

Before we dive into the solution, let’s make sure we fully understand the problem. We are asked to find the limit of the function ${ f(x) = \frac{\sqrt[n]{1+x} - 1}{x} }$ as ${ x }$ approaches 0. This can be written mathematically as:

${ \lim_{x \to 0} \frac{\sqrt[n]{1+x} - 1}{x} }$

Here, ${ n }$ is assumed to be a positive integer. The expression involves a radical term, specifically the nth root of (1+x), which makes the limit a bit challenging to evaluate directly. A direct substitution of ${ x = 0 }$ into the expression results in an indeterminate form of type 0/0, which means we need to employ techniques such as algebraic manipulation or L'Hôpital's Rule to find the limit. The presence of the nth root suggests that we might need to use methods to rationalize the numerator or apply binomial expansion to simplify the expression. The goal is to transform the function into a form where the limit can be easily evaluated. This problem showcases the importance of understanding various limit evaluation techniques and recognizing which method is most appropriate for a given situation. Properly interpreting the limit notation and understanding the behavior of the function near the point of approach are key steps in solving such problems. The following sections will detail the step-by-step solutions using different approaches.

Method 1: Algebraic Manipulation

One way to approach this limit is through algebraic manipulation. The goal is to eliminate the indeterminate form by rationalizing the numerator. To do this, we can multiply the numerator and denominator by a factor that will eliminate the radical in the numerator. Let's consider the expression:

${ \frac{\sqrt[n]{1+x} - 1}{x} }$

To rationalize the numerator, we can multiply both the numerator and the denominator by the conjugate expression. In this case, we need to use a generalized conjugate that will eliminate the nth root. This can be achieved by considering the identity:

${ a^n - b^n = (a - b)(a^{n-1} + a^{n-2}b + a^{n-3}b^2 + \cdots + ab^{n-2} + b^{n-1}) }$

In our case, let ${ a = \sqrt[n]{1+x} }$ and ${ b = 1 }$. Then, we have:

${ a^n - b^n = (1+x) - 1 = x }$

So, we multiply the numerator and denominator by:

${ a^{n-1} + a^{n-2}b + a^{n-3}b^2 + \cdots + ab^{n-2} + b^{n-1} }$

which translates to:

${ (1+x)^{\frac{n-1}{n}} + (1+x)^{\frac{n-2}{n}} + \cdots + (1+x)^{\frac{1}{n}} + 1 }$

Thus, the expression becomes:

${ \lim_{x \to 0} \frac{\sqrt[n]{1+x} - 1}{x} \cdot \frac{(1+x)^{\frac{n-1}{n}} + (1+x)^{\frac{n-2}{n}} + \cdots + (1+x)^{\frac{1}{n}} + 1}{(1+x)^{\frac{n-1}{n}} + (1+x)^{\frac{n-2}{n}} + \cdots + (1+x)^{\frac{1}{n}} + 1} }$

Multiplying the numerator, we get ${ x }$, and the expression simplifies to:

${ \lim_{x \to 0} \frac{x}{x \left[ (1+x)^{\frac{n-1}{n}} + (1+x)^{\frac{n-2}{n}} + \cdots + (1+x)^{\frac{1}{n}} + 1 \right]} }$

We can cancel out the ${ x }$ terms:

${ \lim_{x \to 0} \frac{1}{(1+x)^{\frac{n-1}{n}} + (1+x)^{\frac{n-2}{n}} + \cdots + (1+x)^{\frac{1}{n}} + 1} }$

Now, as ${ x \to 0 }$, each term ${ (1+x)^{\frac{k}{n}} }$ approaches 1. There are ${ n }$ terms in the denominator, so the limit becomes:

${ \frac{1}{1 + 1 + \cdots + 1} = \frac{1}{n} }$

Therefore, the limit of the expression as ${ x }$ approaches 0 is ${ \frac{1}{n} }$. This method demonstrates the power of algebraic manipulation in simplifying complex expressions and evaluating limits.

Method 2: L'Hôpital's Rule

Another powerful tool for evaluating limits is L'Hôpital's Rule. This rule is particularly useful when dealing with indeterminate forms such as 0/0 or ∞/∞. L'Hôpital's Rule states that if the limit of ${ \frac{f(x)}{g(x)} }$ as ${ x }$ approaches a value ${ c }$ results in an indeterminate form, and if ${ f }$ and ${ g }$ are differentiable, 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 side exists. In our case, we have the function:

${ f(x) = \frac{\sqrt[n]{1+x} - 1}{x} }$

As ${ x }$ approaches 0, we have the indeterminate form 0/0. Let's define:

${ u(x) = \sqrt[n]{1+x} - 1 = (1+x)^{\frac{1}{n}} - 1 }$

${ v(x) = x }$

Now, we find the derivatives of ${ u(x) }$ and ${ v(x) }$ with respect to ${ x }$:

${ u'(x) = \frac{1}{n}(1+x)^{\frac{1}{n} - 1} }$

${ v'(x) = 1 }$

Applying L'Hôpital's Rule, we get:

${ \lim_{x \to 0} \frac{u(x)}{v(x)} = \lim_{x \to 0} \frac{u'(x)}{v'(x)} = \lim_{x \to 0} \frac{\frac{1}{n}(1+x)^{\frac{1}{n} - 1}}{1} }$

Now, we evaluate the limit as ${ x }$ approaches 0:

${ \lim_{x \to 0} \frac{1}{n}(1+x)^{\frac{1}{n} - 1} = \frac{1}{n}(1+0)^{\frac{1}{n} - 1} = \frac{1}{n}(1)^{\frac{1}{n} - 1} = \frac{1}{n} }$

Thus, using L'Hôpital's Rule, we find that the limit is ${ \frac{1}{n} }$, which matches the result obtained through algebraic manipulation. This method provides a more direct approach for evaluating the limit, especially when dealing with indeterminate forms. L'Hôpital's Rule is a valuable tool in calculus, simplifying complex limit problems into manageable differentiation exercises. Understanding and applying this rule effectively can save significant time and effort in limit evaluations.

Method 3: Using the Definition of the Derivative

Another insightful approach to evaluating this limit involves recognizing its connection to the definition of the derivative. The derivative of a function ${ f(x) }$ at a point ${ a }$ is defined as:

${ f'(a) = \lim_{x \to a} \frac{f(x) - f(a)}{x - a} }$

Our limit expression is:

${ \lim_{x \to 0} \frac{\sqrt[n]{1+x} - 1}{x} }$

We can rewrite this as:

${ \lim_{x \to 0} \frac{(1+x)^{\frac{1}{n}} - 1}{x} }$

Let's consider a function ${ f(x) = (1+x)^{\frac{1}{n}} }$. We want to find the derivative of this function at ${ x = 0 }$. The limit expression now looks like:

${ \lim_{x \to 0} \frac{f(1+x) - f(1)}{x} }$

If we consider the function ${ g(x) = x }$, we can rewrite the limit as:

${ \lim_{x \to 0} \frac{f(1+x) - f(1)}{(1+x) - 1} }$

If we substitute ${ h = 1 + x }$, then as ${ x \to 0 }$, ${ h \to 1 }$. So, our limit becomes:

${ \lim_{h \to 1} \frac{f(h) - f(1)}{h - 1} }$

However, this form is not exactly matching our original function. Let's stick to the original form and identify ${ f(x) = (1+x)^{\frac{1}{n}} }$. We want to evaluate ${ f'(0) }$, so:

${ f'(0) = \lim_{x \to 0} \frac{f(x) - f(0)}{x - 0} }$

In our case, ${ f(0) = (1+0)^{\frac{1}{n}} = 1 }$. Thus, we have:

${ f'(0) = \lim_{x \to 0} \frac{(1+x)^{\frac{1}{n}} - 1}{x} }$

Now, we need to find the derivative of ${ f(x) = (1+x)^{\frac{1}{n}} }$:

${ f'(x) = \frac{1}{n}(1+x)^{\frac{1}{n} - 1} }$

Evaluating ${ f'(0) }$, we get:

${ f'(0) = \frac{1}{n}(1+0)^{\frac{1}{n} - 1} = \frac{1}{n} }$

Therefore, the limit is ${ \frac{1}{n} }$. This method highlights the connection between limits and derivatives, providing a deeper understanding of calculus concepts. By recognizing the limit as the derivative of a function at a point, we can use differentiation techniques to evaluate the limit directly. This approach is particularly useful for students looking to solidify their understanding of calculus principles and apply them in various problem-solving contexts.

Conclusion

In conclusion, we have successfully evaluated the limit of the expression ${ \frac{\sqrt[n]{1+x} - 1}{x} }$ as ${ x }$ approaches 0 using three different methods: algebraic manipulation, L'Hôpital's Rule, and the definition of the derivative. Each method provides a unique perspective and reinforces the fundamental concepts of calculus. The algebraic manipulation involved rationalizing the numerator using a generalized conjugate, which allowed us to simplify the expression and directly evaluate the limit. L'Hôpital's Rule offered a more direct approach by differentiating the numerator and the denominator and then evaluating the limit of the resulting expression. Finally, recognizing the limit as the derivative of the function ${ f(x) = (1+x)^{\frac{1}{n}} }$ at ${ x = 0 }$ provided an elegant solution through differentiation. All three methods consistently yielded the result of ${ \frac{1}{n} }$. This exercise underscores the importance of having a versatile toolkit of techniques for evaluating limits. Understanding these methods not only helps in solving specific problems but also enhances the comprehension of calculus principles more broadly. The ability to approach a problem from different angles is a hallmark of a strong mathematical thinker, and this example effectively illustrates the power of such flexibility. Whether you are a student learning calculus or a professional applying these concepts, mastering limit evaluation is crucial for success in various fields of study and application.