To prove that

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

exists and equals $e$, we can proceed as follows:

Solution:

1. Consider the expression: $y = \left( 1 + \frac{1}{x} \right)^x$

2. Take the natural logarithm on both sides: $\ln(y) = x \cdot \ln\left(1 + \frac{1}{x}\right)$

3. Analyze the limit of $\ln(y)$ as $x \to \infty$: We want to find: $\lim_{x \to \infty} \ln(y) = \lim_{x \to \infty} x \cdot \ln\left(1 + \frac{1}{x}\right)$

4. Apply Taylor expansion (approximation) for $\ln(1 + u)$ around $u = 0$: Since $\frac{1}{x} \to 0$ as $x \to \infty$, we can use the approximation $\ln(1 + u) \approx u$ when $u$ is small. Therefore: $\ln\left(1 + \frac{1}{x}\right) \approx \frac{1}{x}$

5. Substitute this approximation into the limit: $\lim_{x \to \infty} x \cdot \ln\left(1 + \frac{1}{x}\right) \approx \lim_{x \to \infty} x \cdot \frac{1}{x} = \lim_{x \to \infty} 1 = 1$

6. Interpret this result in terms of $y$: Since $\lim_{x \to \infty} \ln(y) = 1$, we have: $\lim_{x \to \infty} y = e$

Conclusion:

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

This completes the proof that this limit exists and equals $e$.

Would you like more details on this proof or have any specific questions?

Additional Practice Problems

1. Use a similar approach to prove that $\lim_{x \to \infty} \left(1 - \frac{1}{x}\right)^x = \frac{1}{e}$.
2. Explain why the approximation $\ln(1 + u) \approx u$ is valid when $u$ is close to zero.
3. Prove that $\lim_{x \to \infty} \left(1 + \frac{k}{x}\right)^x = e^k$ for any constant $k$.
4. Find the limit $\lim_{x \to \infty} \left(1 + \frac{2}{x}\right)^x$.
5. Investigate the behavior of $\left(1 + \frac{1}{n}\right)^n$ as $n$ approaches infinity numerically.

Tip

When using Taylor approximations, remember that they are most accurate when the input variable approaches the value where the expansion is centered.