Evaluating The Limit Lim N→∞ Logn(∫01(1-x3)ndx) A Comprehensive Guide

Hey guys! Ever stumbled upon a limit that just makes you scratch your head? Well, I recently tackled one that felt like a real brain-teaser:

$$\lim_{n\to \infty}\log_{n}\left(\int_{0}^{1}(1-x^{3})^{n}dx\right)$$

This little gem popped up in a MIT BEE exam, and let me tell you, it's a fantastic example of how calculus and limits can intertwine to create some seriously interesting problems. So, let's break it down, explore my initial attempts, and then dive into a more robust solution. Buckle up, it's going to be a fun ride!

My Initial Approach: A Trigonometric Tango

When I first saw this limit, my brain immediately went into simplification mode. The integral inside the logarithm looked a bit intimidating, especially with that pesky exponent n. My initial thought was to try a trigonometric substitution. Why? Well, expressions like ${1-x^2}$ often play nicely with trig, and ${1-x^3}$ felt like it might have a similar vibe.

I envisioned a substitution along the lines of ${x = \sin^{2/3}(\theta)}$, which would transform ${x^3}$ into ${\sin^2(\theta)}$. This seemed promising because it would turn the integrand into ${(1-\sin^2(\theta))^n = (\cos^2(\theta))^n = \cos^{2n}(\theta)}$, which looks much more manageable.

However, after performing the substitution and working through the details (which involved finding the new limits of integration and dealing with the ${dx}$ term), I hit a bit of a wall. The integral became something like this:

$$\int_{0}^{\pi/2} \cos^{2n}(\theta) \cdot \frac{2}{3} \sin^{-1/3}(\theta) \cos^{-1/3}(\theta) d\theta$$

While the ${\cos^{2n}(\theta)}$ part was indeed simpler, the presence of ${\sin^{-1/3}(\theta)}$ and ${\cos^{-1/3}(\theta)}$ threw a wrench in my plans. I couldn't find a straightforward way to evaluate this integral or to effectively use the limit as ${n\to \infty}$. It felt like I had taken a detour down a trigonometric dead end. I realized I needed a different strategy, one that could directly address the behavior of the integral as n grows large.

Why This Approach Didn't Quite Work

The trigonometric substitution, while a reasonable first attempt, ultimately failed because it introduced new complexities into the integral. The fractional powers of sine and cosine made it difficult to find a closed-form solution or to easily analyze the limit. This highlights an important lesson in problem-solving: sometimes, the most obvious approach isn't the most effective.

I needed a method that could exploit the large n in the exponent of ${(1-x^3)^n}$. This suggested that techniques involving bounding the integral or using Laplace's method might be more fruitful. The key is to recognize how the integrand behaves as n becomes extremely large. The term ${(1-x^3)^n}$ will rapidly approach zero for any ${x > 0}$, meaning the integral's value will be heavily influenced by the behavior of the integrand near ${x = 0}$.

A More Effective Strategy: Bounding the Integral and Squeeze Theorem

Okay, so the trigonometric route didn't pan out. Time to regroup and rethink! The key to this problem, I realized, lies in understanding how the function ${(1-x^3)^n}$ behaves as n approaches infinity. For any ${x}$ strictly between 0 and 1, ${x^3}$ is also between 0 and 1, so ${1-x^3}$ is a positive fraction. Raising a positive fraction to a large power makes it incredibly small. This suggests that the integral's value is mainly determined by what happens near ${x = 0}$.

The Core Idea: Focusing on the Behavior Near x = 0

Near ${x = 0}$, we can use the approximation ${1-x^3 \approx e^{-x^3}}$. This is because the Taylor series expansion of ${e^u}$ around ${u=0}$ is ${1 + u + \frac{u^2}{2!} + ...}$, so for small ${u}$, ${e^u \approx 1 + u}$. Letting ${u = -x^3}$, we get ${e^{-x^3} \approx 1 - x^3}$. This approximation becomes more accurate as ${x}$ gets closer to 0.

Using this approximation, we can rewrite the integral as:

$$\int_{0}^{1} (1-x^3)^n dx \approx \int_{0}^{1} e^{-nx^3} dx$$

This new integral looks much more manageable! Now, let's make a substitution to simplify it further. Let ${u = n^{1/3}x}$, so ${x = \frac{u}{n^{1/3}}}$ and ${dx = \frac{du}{n^{1/3}}}$. The limits of integration remain 0 and ${n^{1/3}}$.

Our integral now transforms into:

$$\int_{0}^{n^{1/3}} e^{-u^3} \frac{du}{n^{1/3}} = \frac{1}{n^{1/3}} \int_{0}^{n^{1/3}} e^{-u^3} du$$

As ${n \to \infty}$, the upper limit of integration, ${n^{1/3}}$, also approaches infinity. We now have:

$$\lim_{n \to \infty} \frac{1}{n^{1/3}} \int_{0}^{\infty} e^{-u^3} du$$

The integral ${\int_{0}^{\infty} e^{-u^3} du}$ is a constant! It's a special case of the Gamma function, specifically ${\frac{1}{3}\Gamma(\frac{1}{3})}$, but we don't need to know the exact value. Let's call this constant ${C}$. So, our integral behaves like ${\frac{C}{n^{1/3}}}$ as ${n}$ gets large.

Applying the Squeeze Theorem

To make this rigorous, we need to show that this approximation is valid. We can do this by bounding the original integral. For ${0 \le x \le 1}$, we have the inequality ${1 - x^3 \le e^{-x^3}}$. This gives us:

$$\int_{0}^{1} (1-x^3)^n dx \le \int_{0}^{1} e^{-nx^3} dx$$

On the other hand, for small ${x}$, we can also find a lower bound. For example, we can use the inequality ${1 - x^3 \ge e^{-2x^3}}$ for sufficiently small ${x}$. This will give us a lower bound on the integral as well. By carefully choosing these bounds and using the Squeeze Theorem, we can rigorously show that:

$$\int_{0}^{1} (1-x^3)^n dx \sim \frac{C}{n^{1/3}}$$

where ${\sim}$ means