To prove that

$\ln(a) = \lim_{n \to \infty} n(a^{1/n} - 1)$

we’ll start by examining the behavior of $a^{1/n}$ as $n \to \infty$ and linking it to the natural logarithm function.

Step 1: Rewrite $a^{1/n}$ in terms of an exponential

Consider the expression $a^{1/n}$: $a^{1/n} = e^{\frac{\ln(a)}{n}}$

Step 2: Expand $e^{\frac{\ln(a)}{n}}$ using the Taylor series

For small values of $x$, we know the Taylor expansion of $e^x$ around $x = 0$ is: $e^x \approx 1 + x$ Applying this to our term $e^{\frac{\ln(a)}{n}}$, we get: $a^{1/n} = e^{\frac{\ln(a)}{n}} \approx 1 + \frac{\ln(a)}{n} \text{ for large } n.$

Step 3: Substitute into $n(a^{1/n} - 1)$

Now, consider the expression $n(a^{1/n} - 1)$: $n(a^{1/n} - 1) \approx n\left(1 + \frac{\ln(a)}{n} - 1\right)$ Simplifying this gives: $n(a^{1/n} - 1) \approx n \cdot \frac{\ln(a)}{n} = \ln(a)$

Step 4: Take the limit as $n \to \infty$

Thus, $\lim_{n \to \infty} n(a^{1/n} - 1) = \ln(a)$

Conclusion

We have shown that: $\ln(a) = \lim_{n \to \infty} n(a^{1/n} - 1)$

This completes the proof.

Would you like further details or have any questions?

Here are some related questions for further exploration:

1. How does the Taylor expansion for $e^x$ work in other approximations?
2. Can this proof be extended to complex numbers?
3. What other ways can we express the natural logarithm in terms of limits?
4. How does this proof connect with the derivative definition of the natural logarithm?
5. Could this limit definition help approximate $\ln(a)$ for small values of $n$?

Tip: Taylor expansions are powerful tools for approximations, especially when dealing with limits involving exponential and logarithmic functions.