Evaluating The Limit Of (1 + 2/x^2)^(x^2) As X Approaches Infinity

#Introduction

In the realm of calculus, evaluating limits is a fundamental skill. This article delves into the process of evaluating a specific limit: lim (x→∞) (1 + 2/x²⁾(x^2). This limit problem is interesting because it involves a function raised to a power, and as x approaches infinity, both the base and the exponent change. To tackle this, we'll employ a strategy that involves recognizing a familiar limit form and using the properties of logarithms and exponentials. This exploration is not just an exercise in calculus; it’s a journey into understanding the behavior of functions as they approach extreme values.

The limit we are tasked with evaluating is:

lim (x→∞) (1 + 2/x^2)^(x^2)

This expression takes the form of a function raised to a power, where both the function and the power depend on x. As x grows infinitely large, the term 2/x^2 approaches zero, making the base (1 + 2/x^2) approach 1. Simultaneously, the exponent x^2 grows without bound. This situation creates an indeterminate form of the type 1^∞, which requires careful analysis to resolve. Indeterminate forms are expressions in calculus where the limit cannot be determined solely from the limits of the individual parts. Other common indeterminate forms include 0/0, ∞/∞, 0 * ∞, ∞ - ∞, 0^0, and ∞^0. Each of these forms necessitates a specific approach to evaluate the limit, often involving techniques such as L'Hôpital's Rule, algebraic manipulation, or, as we will see in this case, the use of exponential and logarithmic transformations.

Understanding how to handle these indeterminate forms is crucial in calculus as they appear frequently in various contexts, including the study of convergence of sequences and series, the analysis of asymptotes, and optimization problems. In this article, we will focus on transforming the 1^∞ form into a more manageable one using the properties of logarithms and exponentials, which will allow us to apply standard limit techniques effectively. This method not only helps in solving this particular problem but also provides a general strategy for dealing with similar limit problems.

Method 1: Using the Exponential and Logarithmic Transformation

To effectively evaluate the limit, we can employ a powerful technique that involves transforming the expression using exponential and logarithmic functions. This method is particularly useful when dealing with indeterminate forms like 1^∞, as it allows us to bring the exponent down and manipulate the expression into a form where L'Hôpital's Rule, or other limit evaluation techniques, can be applied more easily. The core idea behind this transformation is the identity: f(x)^g(x) = e^(g(x) * ln(f(x))), where f(x) is the base of the power and g(x) is the exponent. By applying this identity, we convert the power function into an exponential function, which is often easier to handle in the context of limits. This is because the exponential function is continuous, and we can move the limit inside the exponential, focusing on evaluating the limit of the exponent.

In our specific problem, the limit is:

lim (x→∞) (1 + 2/x^2)^(x^2)

Let's denote the expression inside the limit as y:

y = (1 + 2/x^2)^(x^2)

Now, we apply the exponential and logarithmic transformation by taking the natural logarithm of both sides:

ln(y) = ln((1 + 2/x^2)^(x^2))

Using the power rule of logarithms, which states that ln(a^b) = b * ln(a), we can rewrite the expression as:

ln(y) = x^2 * ln(1 + 2/x^2)

Now, we have transformed the original expression into a product, which is often easier to analyze. The next step is to evaluate the limit of ln(y) as x approaches infinity:

lim (x→∞) ln(y) = lim (x→∞) x^2 * ln(1 + 2/x^2)

As x approaches infinity, the term 2/x^2 approaches 0. Thus, inside the logarithm, we have 1 plus a quantity approaching zero, and outside, we have x^2 approaching infinity. This leads to an indeterminate form of the type ∞ * 0. To resolve this, we can rewrite the product as a fraction, which will allow us to apply L'Hôpital's Rule, a powerful tool for evaluating limits of indeterminate forms of the type 0/0 or ∞/∞. The key to using L'Hôpital's Rule is to differentiate the numerator and the denominator separately and then take the limit. This process can simplify the expression and lead to the determination of the limit. In the following sections, we will detail how to rewrite the expression as a fraction and apply L'Hôpital's Rule to find the limit of ln(y).

Rewriting as a Fraction and Applying L'Hôpital's Rule

In order to apply L'Hôpital's Rule, it is essential to rewrite the limit as a fraction. Recall that we have:

lim (x→∞) ln(y) = lim (x→∞) x^2 * ln(1 + 2/x^2)

This is an indeterminate form of type ∞ * 0. To convert it into a fraction, we can rewrite x^2 as 1/(1/x^2). This gives us:

lim (x→∞) ln(y) = lim (x→∞) ln(1 + 2/x^2) / (1/x^2)

Now, as x approaches infinity, 2/x^2 approaches 0, so ln(1 + 2/x^2) approaches ln(1), which is 0. Similarly, 1/x^2 approaches 0 as x approaches infinity. Thus, we now have an indeterminate form of type 0/0, which is suitable for applying L'Hôpital's Rule.

L'Hôpital's Rule states that if lim (x→c) f(x)/g(x) is of the form 0/0 or ∞/∞, and if the limit of f'(x)/g'(x) exists, then:

lim (x→c) f(x)/g(x) = lim (x→c) f'(x)/g'(x)

Here, f(x) = ln(1 + 2/x^2) and g(x) = 1/x^2. We need to find the derivatives of f(x) and g(x) with respect to x.

First, let's find the derivative of f(x):

f(x) = ln(1 + 2/x^2)

Using the chain rule, we have:

f'(x) = [1 / (1 + 2/x^2)] * d/dx (1 + 2/x^2)

The derivative of 1 + 2/x^2 is -4/x^3. So,

f'(x) = [1 / (1 + 2/x^2)] * (-4/x^3)

Now, let's find the derivative of g(x):

g(x) = 1/x^2 = x^(-2)

g'(x) = -2x^(-3) = -2/x^3

Now we can apply L'Hôpital's Rule:

lim (x→∞) ln(y) = lim (x→∞) f'(x)/g'(x)

lim (x→∞) ln(y) = lim (x→∞)  [ (1 / (1 + 2/x^2)) * (-4/x^3) ] / (-2/x^3)

We can simplify this expression by canceling out the -4/x^3 and -2/x^3 terms:

lim (x→∞) ln(y) = lim (x→∞) 2 / (1 + 2/x^2)

As x approaches infinity, 2/x^2 approaches 0, so we have:

lim (x→∞) ln(y) = 2 / (1 + 0) = 2

Therefore, the limit of ln(y) as x approaches infinity is 2. However, we need to find the limit of y, not ln(y). To do this, we use the fact that if lim (x→c) ln(y) = L, then lim (x→c) y = e^L. This property is a direct consequence of the continuity of the exponential function.

Finding the Original Limit

After applying L'Hôpital's Rule, we found that:

lim (x→∞) ln(y) = 2

where y = (1 + 2/x²⁾(x^2). To find the original limit, which is lim (x→∞) y, we need to