Evaluate the following limits :
`lim_(x to 0)[1/x-(log(1+x))/x^(2)]`

Answer

To evaluate the limit

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

we start by substituting \( x = 0 \):

1. **Substitution**:
\[ \frac{1}{0} - \frac{\log(1+0)}{0^2} = \frac{1}{0} - \frac{0}{0} \text{ (which is an indeterminate form)} \]

Since we have an indeterminate form \( \frac{0}{0} \), we can apply L'Hôpital's Rule.

2. **Apply L'Hôpital's Rule**:
We differentiate the numerator and the denominator separately.

The expression can be rewritten as:
\[ \lim_{x \to 0} \frac{\log(1+x) - x}{x^2}. \]

Now we differentiate the numerator and denominator:
- The derivative of the numerator \( \log(1+x) - x \) is:
\[ \frac{1}{1+x} - 1 = \frac{1 - (1+x)}{1+x} = \frac{-x}{1+x}. \]
- The derivative of the denominator \( x^2 \) is:
\[ 2x. \]

Thus, we have:
\[ \lim_{x \to 0} \frac{-x/(1+x)}{2x}. \]

3. **Simplify the expression**:
\[ = \lim_{x \to 0} \frac{-1}{2(1+x)}. \]

4. **Evaluate the limit**:
Now we can substitute \( x = 0 \):
\[ = \frac{-1}{2(1+0)} = \frac{-1}{2}. \]

Therefore, the final answer is:
\[ \lim_{x \to 0} \left( \frac{1}{x} - \frac{\log(1+x)}{x^2} \right) = -\frac{1}{2}. \]