`lim_(x to 0) [1/x-1/x^(2) log (1+x)]` =

Options

A. 1
B. `1//2`
C. `-1`
D. none

Answer

To solve the limit

\[ \lim_{x \to 0} \left( \frac{1}{x} - \frac{1}{x^2} \log(1+x) \right), \]

we will first expand \(\log(1+x)\) using its Taylor series expansion around \(x = 0\):

\[ \log(1+x) = x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + \ldots \]

Now substituting this expansion into our limit expression:

\[ \lim_{x \to 0} \left( \frac{1}{x} - \frac{1}{x^2} \left( x - \frac{x^2}{2} + \frac{x^3}{3} - \ldots \right) \right) \]

Distributing \(-\frac{1}{x^2}\):

\[ = \lim_{x \to 0} \left( \frac{1}{x} - \frac{x}{x^2} + \frac{1}{2} - \frac{x}{3} + \ldots \right) \]

This simplifies to:

\[ = \lim_{x \to 0} \left( \frac{1}{x} - \frac{1}{x} + \frac{1}{2} - \frac{x}{3} + \ldots \right) \]

Notice that the \(\frac{1}{x}\) terms cancel out:

\[ = \lim_{x \to 0} \left( \frac{1}{2} - \frac{x}{3} + \ldots \right) \]

As \(x\) approaches \(0\), all terms involving \(x\) will vanish:

\[ = \frac{1}{2} \]

Thus, the limit evaluates to:

\[ \boxed{\frac{1}{2}} \]