`lim_(n to infty) [1- log(1+1/n)^(n-1)]` is…………

Answer

To solve the limit \( \lim_{n \to \infty} \left[ 1 - \log\left(1 + \frac{1}{n}\right)^{n-1} \right] \), we can follow these steps:

### Step 1: Rewrite the logarithmic expression
Using the property of logarithms, we can rewrite the expression inside the limit:
\[ \log\left(1 + \frac{1}{n}\right)^{n-1} = (n-1) \log\left(1 + \frac{1}{n}\right) \]
Thus, the limit becomes:
\[ \lim_{n \to \infty} \left[ 1 - (n-1) \log\left(1 + \frac{1}{n}\right) \right] \]

### Step 2: Expand the logarithm using Taylor series
For small values of \( x \), we can use the Taylor series expansion:
\[ \log(1 + x) \approx x - \frac{x^2}{2} + \frac{x^3}{3} - \ldots \]
Substituting \( x = \frac{1}{n} \):
\[ \log\left(1 + \frac{1}{n}\right) \approx \frac{1}{n} - \frac{1}{2n^2} + O\left(\frac{1}{n^3}\right) \]

### Step 3: Substitute the expansion back into the limit
Now substituting this expansion back into our limit:
\[ (n-1) \log\left(1 + \frac{1}{n}\right) \approx (n-1)\left(\frac{1}{n} - \frac{1}{2n^2} + O\left(\frac{1}{n^3}\right)\right) \]
This simplifies to:
\[ (n-1)\left(\frac{1}{n}\right) - (n-1)\left(\frac{1}{2n^2}\right) + O\left(\frac{n-1}{n^3}\right) \]
\[ = 1 - \frac{1}{n} - \frac{1}{2n} + O\left(\frac{1}{n^2}\right) \]

### Step 4: Substitute back into the limit expression
Now substituting this back into our limit:
\[ \lim_{n \to \infty} \left[ 1 - \left(1 - \frac{1}{n} - \frac{1}{2n} + O\left(\frac{1}{n^2}\right)\right) \right] \]
This simplifies to:
\[ \lim_{n \to \infty} \left[ \frac{1}{n} + \frac{1}{2n} - O\left(\frac{1}{n^2}\right) \right] \]
\[ = \lim_{n \to \infty} \left[ \frac{3}{2n} - O\left(\frac{1}{n^2}\right) \right] \]

### Step 5: Evaluate the limit
As \( n \to \infty \), both terms \( \frac{3}{2n} \) and \( O\left(\frac{1}{n^2}\right) \) approach 0:
\[ \lim_{n \to \infty} \left[ \frac{3}{2n} \right] = 0 \]

Thus, the final result is:
\[ \lim_{n \to \infty} \left[ 1 - \log\left(1 + \frac{1}{n}\right)^{n-1} \right] = 0 \]

### Final Answer:
The limit is \( 0 \).