Answer:
To calculate the limit as x approaches infinity of the given expression:
\[ \lim_{x\to\infty} \frac{x}{x - \ln(1+x)} \]
We can use the series expansion for \(\ln(1+x)\) that you provided:
\[ \ln(1+x) = \sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n} x^n \]
Now, let's substitute this series into the original expression:
\[ \lim_{x\to\infty} \frac{x}{x - \sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n} x^n} \]
Now, let's take a look at the denominator, which is a series. We can rewrite it as:
\[ \sum_{n=1}^{\infty} \left(\frac{(-1)^{n-1}}{n} x^n\right) = x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + \ldots \]
As \(x\) approaches infinity, all the terms with \(x^n\) where \(n\) is a positive integer become much larger than the terms with smaller exponents. So, we can approximate the denominator as:
\[ x - \frac{x^2}{2} \]
Now, let's go back to the original expression:
\[ \lim_{x\to\infty} \frac{x}{x - \sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n} x^n} = \lim_{x\to\infty} \frac{x}{x - \left(x - \frac{x^2}{2}\right)} \]
Simplify the expression:
\[ \lim_{x\to\infty} \frac{x}{\frac{x^2}{2}} \]
Now, divide both the numerator and denominator by \(x\):
\[ \lim_{x\to\infty} \frac{1}{\frac{1}{2}x} \]
As \(x\) approaches infinity, the limit becomes:
\[ \lim_{x\to\infty} 0 \]
So, the limit as \(x\) approaches infinity of the given expression is 0.
Explanation: