Limit Calculation: Solve The Expression As N Approaches Infinity

by ADMIN 65 views

Hey guys! Today, we're diving deep into a fascinating problem in calculus: calculating the limit of a complex expression as n approaches infinity. This isn't just about plugging in a value; it's about understanding how functions behave when they grow without bound. So, let's break down the problem step by step and make sure we not only get the answer but also grasp the underlying concepts.

The Problem Unveiled

The expression we're tackling is:

lim (n→∞) [(1+1/n)(1+2/n)⋯(1+4n/n)]^(1/n)

At first glance, it might seem intimidating with its product of terms and the exponent. But don't worry! We'll use some clever mathematical tools to simplify it. The key here is to recognize that this is a classic problem that can be elegantly solved using logarithms and the properties of limits.

The Logarithmic Transformation

The first strategic move is to take the natural logarithm (ln) of the entire expression. Why? Because logarithms have a magical property: they turn products into sums and exponents into multipliers. This transformation makes the expression much easier to handle. So, let's define L as the limit we want to find, and apply the natural logarithm:

Let L = lim (n→∞) [(1+1/n)(1+2/n)⋯(1+4n/n)]^(1/n)

ln(L) = ln(lim (n→∞) [(1+1/n)(1+2/n)⋯(1+4n/n)]^(1/n))

Using the properties of logarithms, we can rewrite this as:

ln(L) = lim (n→∞) ln([(1+1/n)(1+2/n)⋯(1+4n/n)]^(1/n))

ln(L) = lim (n→∞) (1/n) * ln[(1+1/n)(1+2/n)⋯(1+4n/n)]

Now, we can use another logarithmic property to expand the logarithm of the product into a sum of logarithms:

ln(L) = lim (n→∞) (1/n) * [ln(1+1/n) + ln(1+2/n) + ⋯ + ln(1+4n/n)]

See how the expression is already looking much friendlier? We've turned a product inside a logarithm into a sum of logarithms, which is a significant step forward.

Recognizing the Riemann Sum

Now comes the brilliant part! We need to recognize that the sum inside the limit looks a lot like a Riemann sum. A Riemann sum is a way to approximate the definite integral of a function, and it's defined as the sum of the areas of rectangles under the curve. To make the connection clearer, let's rewrite the sum using sigma notation:

ln(L) = lim (n→∞) (1/n) * Σ [ln(1 + k/n)] (where k ranges from 1 to 4n)

This form is crucial because it directly corresponds to the definition of a Riemann sum. We can see that:

  • 1/n represents the width of each rectangle.
  • ln(1 + k/n) represents the height of each rectangle, which is the value of the function ln(1 + x) at the point x = k/n.
  • The summation Σ represents the sum of the areas of all the rectangles.

As n approaches infinity, the width of the rectangles (1/n) approaches zero, and the Riemann sum converges to the definite integral of the function. This is a fundamental concept in calculus, and it's the key to solving our problem.

Converting to a Definite Integral

With the Riemann sum identified, we can now convert the limit of the sum into a definite integral. The interval of integration is determined by the limits of the summation. Here, k ranges from 1 to 4n, so k/n ranges from 1/n to 4. As n approaches infinity, 1/n approaches 0, so our interval of integration is [0, 4].

The function we're integrating is ln(1 + x), which we identified from the height of the rectangles in the Riemann sum. Therefore, we can rewrite the limit as a definite integral:

ln(L) = ∫[0 to 4] ln(1 + x) dx

Now, we've transformed our original limit problem into a definite integral, which is a much more manageable form. We can use integration techniques to evaluate this integral and find the value of ln(L).

Evaluating the Definite Integral

To evaluate the definite integral ∫[0 to 4] ln(1 + x) dx, we'll use integration by parts. Integration by parts is a technique that helps us integrate the product of two functions. The formula for integration by parts is:

∫ u dv = uv - ∫ v du

In our case, we'll choose:

  • u = ln(1 + x)
  • dv = dx

Then, we need to find du and v:

  • du = (1/(1 + x)) dx
  • v = x

Now, we can apply the integration by parts formula:

∫[0 to 4] ln(1 + x) dx = [x * ln(1 + x)][from 0 to 4] - ∫[0 to 4] x * (1/(1 + x)) dx

Let's evaluate the first term and simplify the integral:

[x * ln(1 + x)][from 0 to 4] = 4 * ln(5) - 0 * ln(1) = 4 * ln(5)

Now, we need to evaluate the remaining integral:

∫[0 to 4] x/(1 + x) dx

To do this, we can use a little trick: add and subtract 1 in the numerator:

∫[0 to 4] (x + 1 - 1)/(1 + x) dx = ∫[0 to 4] (1 - 1/(1 + x)) dx

Now, we can integrate each term separately:

∫[0 to 4] (1 - 1/(1 + x)) dx = [x - ln(1 + x)][from 0 to 4]

Evaluate the expression at the limits of integration:

[4 - ln(5)] - [0 - ln(1)] = 4 - ln(5)

Now, we can substitute these results back into our integration by parts equation:

∫[0 to 4] ln(1 + x) dx = 4 * ln(5) - (4 - ln(5))

Simplify the expression:

∫[0 to 4] ln(1 + x) dx = 4 * ln(5) - 4 + ln(5) = 5 * ln(5) - 4

So, we've finally evaluated the definite integral! We found that:

∫[0 to 4] ln(1 + x) dx = 5 * ln(5) - 4

Solving for L

Remember that we took the natural logarithm of L at the beginning of our journey. Now, we need to reverse that process to find the value of L itself. We know that:

ln(L) = 5 * ln(5) - 4

To get L, we need to exponentiate both sides of the equation using the exponential function (e^x):

L = e^(5 * ln(5) - 4)

Using the properties of exponents, we can rewrite this as:

L = e^(5 * ln(5)) * e^(-4)

We can further simplify using the property e^(ln(x)) = x:

L = (e(ln(5)))5 * e^(-4)

L = 5^5 * e^(-4)

L = 5^5 / e^4

The Grand Finale: The Answer

After all that mathematical maneuvering, we've arrived at the solution! The limit of the expression as n approaches infinity is:

L = 5^5 / e^4

So, the correct answer is (c) 5^5 / e^4.

Key Takeaways

This problem was a fantastic exercise in using several important calculus concepts:

  • Logarithmic Transformation: Taking the logarithm of a complex expression can simplify it significantly, especially when dealing with products and exponents.
  • Riemann Sums: Recognizing Riemann sums allows us to convert limits of sums into definite integrals, a powerful technique in calculus.
  • Integration by Parts: This technique is essential for integrating products of functions and was crucial for evaluating our definite integral.
  • Properties of Limits and Exponents: Understanding these properties is fundamental for manipulating and simplifying expressions.

Final Thoughts

Solving problems like this can seem daunting at first, but by breaking them down into smaller steps and applying the right techniques, we can conquer even the most complex mathematical challenges. The key is to practice, understand the underlying concepts, and never be afraid to try different approaches.

Keep exploring the fascinating world of calculus, guys! There's always something new and exciting to learn. And remember, math isn't just about finding the right answer; it's about the journey of discovery and the joy of understanding.