The value of `lim_(xto0)(x cosx-log(1+x))/(x^(2))` is

Options

A. `1`
B. `1/4`
C. `1/2`
D. None of these

Answer

To solve the limit \( \lim_{x \to 0} \frac{x \cos x - \log(1+x)}{x^2} \), we can use the Taylor series expansions for \( \cos x \) and \( \log(1+x) \).

### Step-by-step Solution:

1. **Expand \( \cos x \)**:
The Taylor series expansion of \( \cos x \) around \( x = 0 \) is:
\[ \cos x = 1 - \frac{x^2}{2} + \frac{x^4}{24} - \cdots \]

2. **Expand \( \log(1+x) \)**:
The Taylor series expansion of \( \log(1+x) \) around \( x = 0 \) is:
\[ \log(1+x) = x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + \cdots \]

3. **Substitute the expansions into the limit**:
We substitute these expansions into the limit expression:
\[ x \cos x - \log(1+x) = x \left(1 - \frac{x^2}{2} + \frac{x^4}{24} - \cdots\right) - \left(x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + \cdots\right) \]

This simplifies to:
\[ = x - \frac{x^3}{2} + \frac{x^5}{24} - x + \frac{x^2}{2} - \frac{x^3}{3} + \frac{x^4}{4} + \cdots \]

Combining like terms:
\[ = \left(-\frac{x^3}{2} + \frac{x^2}{2} - \frac{x^3}{3} + \cdots\right) \]

4. **Combine the terms**:
The leading terms after simplification will be:
\[ = \frac{x^2}{2} - \left(\frac{3}{6} + \frac{2}{6}\right)x^3 + \cdots \]
The \( x \) terms cancel out, and we focus on the \( x^2 \) and \( x^3 \) terms.

5. **Divide by \( x^2 \)**:
Now we divide the entire expression by \( x^2 \):
\[ \frac{\frac{x^2}{2} - \left(\frac{5}{6}\right)x^3 + \cdots}{x^2} = \frac{1}{2} - \frac{5}{6}x + \cdots \]

6. **Take the limit as \( x \to 0 \)**:
As \( x \) approaches 0, the higher order terms vanish:
\[ \lim_{x \to 0} \left(\frac{1}{2} - \frac{5}{6}x + \cdots\right) = \frac{1}{2} \]

Thus, the value of the limit is:
\[ \boxed{\frac{1}{2}} \]