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 find the limit

\[ L = \lim_{x \to 0} \frac{x \cos x - \log(1+x)}{x^2}, \]

we first observe that substituting \(x = 0\) directly into the expression gives us the indeterminate form \(\frac{0}{0}\). Therefore, we can apply L'Hôpital's Rule, which states that if we have an indeterminate form \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\), we can take the derivative of the numerator and the derivative of the denominator.

### Step 1: Apply L'Hôpital's Rule

We differentiate the numerator and the denominator:

- The numerator is \(x \cos x - \log(1+x)\).
- The derivative of \(x \cos x\) is given by the product rule:
\[ \frac{d}{dx}(x \cos x) = \cos x - x \sin x. \]
- The derivative of \(-\log(1+x)\) is:
\[ \frac{d}{dx}(-\log(1+x)) = -\frac{1}{1+x}. \]

Thus, the derivative of the numerator is:
\[ \cos x - x \sin x - \frac{1}{1+x}. \]

The derivative of the denominator \(x^2\) is:
\[ \frac{d}{dx}(x^2) = 2x. \]

Now we rewrite the limit using these derivatives:
\[ L = \lim_{x \to 0} \frac{\cos x - x \sin x - \frac{1}{1+x}}{2x}. \]

### Step 2: Substitute \(x = 0\) Again

Substituting \(x = 0\) into the new expression:
- \(\cos(0) = 1\),
- \(-0 \cdot \sin(0) = 0\),
- \(-\frac{1}{1+0} = -1\).

Thus, the numerator becomes:
\[ 1 - 0 - 1 = 0. \]

The denominator becomes:
\[ 2 \cdot 0 = 0. \]

Again, we have the indeterminate form \(\frac{0}{0}\). We apply L'Hôpital's Rule again.

### Step 3: Differentiate Again

We differentiate the numerator and denominator again:

- The derivative of the numerator \(\cos x - x \sin x - \frac{1}{1+x}\) is:
\[ -\sin x - (\sin x + x \cos x) + \frac{1}{(1+x)^2}. \]
Simplifying this gives:
\[ -2\sin x - x \cos x + \frac{1}{(1+x)^2}. \]

- The derivative of the denominator \(2x\) is:
\[ 2. \]

Now we rewrite the limit:
\[ L = \lim_{x \to 0} \frac{-2\sin x - x \cos x + \frac{1}{(1+x)^2}}{2}. \]

### Step 4: Substitute \(x = 0\) Again

Substituting \(x = 0\):
- \(-2\sin(0) = 0\),
- \(-0 \cdot \cos(0) = 0\),
- \(\frac{1}{(1+0)^2} = 1\).

Thus, the numerator becomes:
\[ 0 - 0 + 1 = 1. \]

The denominator is \(2\).

### Final Result

Thus, we have:
\[ L = \frac{1}{2}. \]

So, the value of the limit is

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