The problem involves evaluating the following limit:

$\lim_{x \to +\infty} \left(1 - \frac{1}{x^3}\right)^x$

Let's break it down step by step:

Step 1: Take the natural logarithm

Rewrite the expression using the logarithm property to make the power $x$ easier to handle: $y = \left(1 - \frac{1}{x^3}\right)^x \quad \implies \quad \ln y = x \ln\left(1 - \frac{1}{x^3}\right)$

Step 2: Expand $\ln\left(1 - \frac{1}{x^3}\right)$ using a Taylor approximation

For large $x$, $\ln(1 - u) \approx -u$, where $u = \frac{1}{x^3}$. Thus: $\ln\left(1 - \frac{1}{x^3}\right) \approx -\frac{1}{x^3}.$

So: $\ln y \approx x \cdot \left(-\frac{1}{x^3}\right) = -\frac{1}{x^2}.$

Step 3: Exponentiate the result

Exponentiating both sides gives: $y \approx e^{-\frac{1}{x^2}}.$

For $x \to +\infty$, $-\frac{1}{x^2} \to 0$, so: $y \to e^0 = 1.$

Final Answer:

$\lim_{x \to +\infty} \left(1 - \frac{1}{x^3}\right)^x = 1.$

Would you like me to expand further on any step? Let me know! Here are some related questions:

Related Questions:

1. What is the general formula for Taylor series expansion of $\ln(1-u)$?
2. How does the approximation $(1-u)^n \approx e^{-nu}$ for small $u$ work?
3. What are other ways to solve limits involving exponential growth or decay?
4. How would this problem change if the exponent $x$ were replaced by $x^2$?
5. Can you derive the result graphically or numerically?

Tip:

Always check the growth of the base and exponent separately when solving limits of exponential forms!