Evaluating Limit Of ((2^(1/x) + 4^(1/x) + 8^(1/x)) / 3)^x As X Approaches Infinity
Introduction
In the realm of mathematical analysis, evaluating limits is a fundamental concept. Limits help us understand the behavior of functions as their input approaches a specific value, whether it's a finite number or infinity. This article delves into the intricacies of evaluating a particular limit: the limit as x approaches infinity of the expression ((2^(1/x) + 4^(1/x) + 8^(1/x)) / 3 )^x. This problem combines exponential functions, fractional exponents, and the concept of limits at infinity, making it a rich exercise in calculus. Understanding how to solve such limits is crucial for various applications, including understanding the asymptotic behavior of functions, determining convergence of sequences and series, and more. We will explore the necessary steps and techniques to tackle this problem effectively.
Problem Statement
The core of our discussion is the evaluation of the following limit:
lim (x→∞) ((2^(1/x) + 4^(1/x) + 8^(1/x)) / 3 )^x
This limit presents a unique challenge due to its form. As x approaches infinity, 1/x approaches 0. This causes the terms 2^(1/x), 4^(1/x), and 8^(1/x) to approach 1. Therefore, the expression inside the parentheses approaches (1 + 1 + 1) / 3 = 1. However, the exponent x approaches infinity, leading to an indeterminate form of the type 1^∞. Indeterminate forms necessitate careful analysis and cannot be evaluated by direct substitution. The 1^∞ form often requires the use of techniques such as L'Hôpital's Rule, logarithmic transformations, or recognizing standard limit forms. To successfully evaluate this limit, we need to transform the expression into a more manageable form. This involves leveraging properties of exponents and logarithms, and potentially applying L'Hôpital's Rule to resolve the indeterminate form. The solution will reveal how the interplay between the base approaching 1 and the exponent approaching infinity determines the final value of the limit.
Method 1: Using Exponential and Logarithmic Transformation
One effective method to tackle limits of the form 1^∞ involves using exponential and logarithmic transformations. This technique allows us to bring the exponent down and transform the indeterminate power into an indeterminate product or quotient, which we can then handle using other methods, such as L'Hôpital's Rule.
Step 1: Transformation
Let's denote the given expression as:
y = ((2^(1/x) + 4^(1/x) + 8^(1/x)) / 3 )^x
To eliminate the exponent, we take the natural logarithm (ln) of both sides:
ln(y) = ln(((2^(1/x) + 4^(1/x) + 8^(1/x)) / 3 )^x)
Using the property of logarithms that ln(a^b) = b * ln(a), we get:
ln(y) = x * ln((2^(1/x) + 4^(1/x) + 8^(1/x)) / 3)
Now, we need to evaluate the limit of ln(y) as x approaches infinity:
lim (x→∞) ln(y) = lim (x→∞) x * ln((2^(1/x) + 4^(1/x) + 8^(1/x)) / 3)
Step 2: Rewriting the Limit
The limit is now in the form ∞ * ln(1), which is an indeterminate form of the type ∞ * 0. To apply L'Hôpital's Rule, we need to rewrite the expression as a fraction. We can do this by rewriting the limit as:
lim (x→∞) ln(y) = lim (x→∞) ln((2^(1/x) + 4^(1/x) + 8^(1/x)) / 3) / (1/x)
This transformation changes the indeterminate form from ∞ * 0 to 0 / 0, which is suitable for L'Hôpital's Rule.
Step 3: Applying L'Hôpital's Rule
L'Hôpital's Rule states that if we have a limit of the form 0 / 0 or ∞ / ∞, we can take the derivatives of the numerator and the denominator separately and then evaluate the limit:
lim (x→∞) f(x) / g(x) = lim (x→∞) f'(x) / g'(x), provided the limit on the right exists.
In our case, f(x) = ln((2^(1/x) + 4^(1/x) + 8^(1/x)) / 3) and g(x) = 1/x. We need to find the derivatives f'(x) and g'(x).
First, let's find g'(x):
g'(x) = d/dx (1/x) = -1/x^2
Next, let's find f'(x). This is a bit more involved due to the chain rule. Let u(x) = (2^(1/x) + 4^(1/x) + 8^(1/x)) / 3. Then f(x) = ln(u(x)), and f'(x) = (1/u(x)) * u'(x).
Now, we need to find u'(x). Let's find the derivative of each term in u(x) separately:
d/dx (2^(1/x)) = 2^(1/x) * ln(2) * (-1/x^2) d/dx (4^(1/x)) = 4^(1/x) * ln(4) * (-1/x^2) d/dx (8^(1/x)) = 8^(1/x) * ln(8) * (-1/x^2)
So,
u'(x) = (1/3) * [-2^(1/x) * ln(2) / x^2 - 4^(1/x) * ln(4) / x^2 - 8^(1/x) * ln(8) / x^2]
Now we can find f'(x):
f'(x) = (1/u(x)) * u'(x)
f'(x) = (3 / (2^(1/x) + 4^(1/x) + 8^(1/x))) * (1/3) * [-2^(1/x) * ln(2) / x^2 - 4^(1/x) * ln(4) / x^2 - 8^(1/x) * ln(8) / x^2]
f'(x) = [-2^(1/x) * ln(2) - 4^(1/x) * ln(4) - 8^(1/x) * ln(8)] / [x^2 * (2^(1/x) + 4^(1/x) + 8^(1/x))]
Now we apply L'Hôpital's Rule:
lim (x→∞) ln(y) = lim (x→∞) f'(x) / g'(x)
lim (x→∞) ln(y) = lim (x→∞) [-2^(1/x) * ln(2) - 4^(1/x) * ln(4) - 8^(1/x) * ln(8)] / [x^2 * (2^(1/x) + 4^(1/x) + 8^(1/x))] / (-1/x^2)
Step 4: Simplifying and Evaluating the Limit
We can simplify the expression by canceling out the x^2 terms:
lim (x→∞) ln(y) = lim (x→∞) [2^(1/x) * ln(2) + 4^(1/x) * ln(4) + 8^(1/x) * ln(8)] / [2^(1/x) + 4^(1/x) + 8^(1/x)]
As x approaches infinity, 1/x approaches 0. Therefore, 2^(1/x), 4^(1/x), and 8^(1/x) all approach 1. Thus, we can substitute these values into the limit:
lim (x→∞) ln(y) = [1 * ln(2) + 1 * ln(4) + 1 * ln(8)] / [1 + 1 + 1]
lim (x→∞) ln(y) = [ln(2) + ln(4) + ln(8)] / 3
Using the property of logarithms that ln(a * b) = ln(a) + ln(b), and noting that 4 = 2^2 and 8 = 2^3, we have ln(4) = 2ln(2) and ln(8) = 3ln(2):
lim (x→∞) ln(y) = [ln(2) + 2ln(2) + 3ln(2)] / 3
lim (x→∞) ln(y) = [6ln(2)] / 3
lim (x→∞) ln(y) = 2ln(2)
Step 5: Finding the Original Limit
We found the limit of ln(y). To find the limit of y, we need to exponentiate both sides:
lim (x→∞) y = e^(lim (x→∞) ln(y))
lim (x→∞) y = e^(2ln(2))
Using the property of exponents that e^(k*ln(a)) = a^k:
lim (x→∞) y = 2^2
lim (x→∞) y = 4
Therefore, the limit as x approaches infinity of ((2^(1/x) + 4^(1/x) + 8^(1/x)) / 3 )^x is 4.
Method 2: Using the Standard Limit Form
Another approach to solving this limit involves recognizing and utilizing a standard limit form. This method is particularly effective when dealing with expressions that can be manipulated to resemble a known limit. The standard limit form we will use here is:
lim (x→0) (1 + x)^(1/x) = e
This limit is a cornerstone in calculus and is closely related to the definition of the exponential function. By creatively manipulating our given expression, we can transform it into a form where we can apply this standard limit.
Step 1: Rewriting the Expression
Let's start by revisiting the original limit:
lim (x→∞) ((2^(1/x) + 4^(1/x) + 8^(1/x)) / 3 )^x
Our goal is to rewrite the expression inside the parentheses in the form of 1 + something. To do this, we subtract 1 and add 1 inside the parentheses:
lim (x→∞) (1 + ((2^(1/x) + 4^(1/x) + 8^(1/x)) / 3 - 1))^x
Now, let's simplify the expression inside the brackets:
lim (x→∞) (1 + (2^(1/x) + 4^(1/x) + 8^(1/x) - 3) / 3)^x
Step 2: Introducing a Substitution
To further simplify and align with the standard limit form, let's make a substitution. Let u = 1/x. As x approaches infinity, u approaches 0. So, we can rewrite the limit in terms of u:
lim (u→0) (1 + (2^u + 4^u + 8^u - 3) / 3)^(1/u)
Now, let's focus on the term (2^u + 4^u + 8^u - 3) / 3. We want to express this term in a way that we can use the standard limit form.
Step 3: Manipulating the Expression
Let's rewrite the expression as:
lim (u→0) (1 + ((2^u - 1) + (4^u - 1) + (8^u - 1)) / 3)^(1/u)
This form is helpful because we know a standard limit for expressions of the form (a^u - 1) / u as u approaches 0:
lim (u→0) (a^u - 1) / u = ln(a)
Step 4: Applying Known Limits
To use this standard limit, we can multiply and divide each term by u:
lim (u→0) (1 + ((2^u - 1) / u * u + (4^u - 1) / u * u + (8^u - 1) / u * u) / 3)^(1/u)
Now, we can factor out u from the numerator:
lim (u→0) (1 + u * ((2^u - 1) / u + (4^u - 1) / u + (8^u - 1) / u) / 3)^(1/u)
Let's define a new function:
v(u) = ((2^u - 1) / u + (4^u - 1) / u + (8^u - 1) / u) / 3
As u approaches 0, we can apply the standard limit lim (u→0) (a^u - 1) / u = ln(a) to each term:
lim (u→0) v(u) = (ln(2) + ln(4) + ln(8)) / 3
We already computed this in the previous method:
lim (u→0) v(u) = (ln(2) + 2ln(2) + 3ln(2)) / 3 = 6ln(2) / 3 = 2ln(2)
Step 5: Transforming into the Standard Form
Now we have:
lim (u→0) (1 + u * v(u))^(1/u)
To match the standard limit form lim (x→0) (1 + x)^(1/x) = e, we need to rewrite the exponent. Let's multiply the exponent by v(u) and divide by v(u):
lim (u→0) (1 + u * v(u))^(1 / (u * v(u)) * v(u))
Now we can rewrite the limit using the property (ab)c = a^(b*c):
lim (u→0) [(1 + u * v(u))^(1 / (u * v(u)))]^(v(u))
As u approaches 0, u * v(u) also approaches 0. Thus, we can apply the standard limit:
lim (u→0) (1 + u * v(u))^(1 / (u * v(u))) = e
Step 6: Final Evaluation
So, our limit becomes:
lim (u→0) e^(v(u)) = e^(lim (u→0) v(u))
We already found that lim (u→0) v(u) = 2ln(2):
e^(2ln(2))
Using the property e^(k*ln(a)) = a^k:
e^(2ln(2)) = 2^2 = 4
Therefore, the limit as x approaches infinity of ((2^(1/x) + 4^(1/x) + 8^(1/x)) / 3 )^x is 4.
Conclusion
In conclusion, we have successfully evaluated the limit as x approaches infinity of ((2^(1/x) + 4^(1/x) + 8^(1/x)) / 3 )^x using two distinct methods. The first method involved logarithmic transformation and the application of L'Hôpital's Rule, while the second method utilized a standard limit form. Both approaches led us to the same result, which is 4. This exercise highlights the importance of recognizing indeterminate forms and the versatility of techniques available for limit evaluation. Mastering these techniques is essential for further studies in calculus and mathematical analysis. The problem also underscores the significance of understanding the behavior of exponential functions and their interplay with limits, particularly at infinity. The consistent result obtained through different methods reinforces the correctness of our approach and provides a deeper understanding of the underlying mathematical principles involved.
