`lim _(xto0) ((cos x -secx )/(x ^(2) (x+1)))=`

Options

A. 0
B. `-1/2`
C. `-1`
D. `-2`

Answer

To solve the limit

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

we first need to analyze the expression. When we substitute \(x = 0\) directly, we get:

\[ \cos(0) - \sec(0) = 1 - 1 = 0, \]

and the denominator becomes:

\[ 0^2(0 + 1) = 0. \]

This results in the indeterminate form \(\frac{0}{0}\). To resolve this, 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: Differentiate the numerator and denominator

The numerator is \( \cos x - \sec x \). The derivative of this is:

\[ \frac{d}{dx}(\cos x - \sec x) = -\sin x - \sec x \tan x. \]

The denominator is \( x^2 (x + 1) \). We can use the product rule to differentiate this:

\[ \frac{d}{dx}(x^2 (x + 1)) = (2x)(x + 1) + (x^2)(1) = 2x^2 + 2x + x^2 = 3x^2 + 2x. \]

### Step 2: Apply L'Hôpital's Rule

Now we can apply L'Hôpital's Rule:

\[ \lim_{x \to 0} \frac{\cos x - \sec x}{x^2 (x + 1)} = \lim_{x \to 0} \frac{-\sin x - \sec x \tan x}{3x^2 + 2x}. \]

### Step 3: Substitute \(x = 0\) again

Substituting \(x = 0\) in the new limit gives:

Numerator:

\[ -\sin(0) - \sec(0) \tan(0) = 0 - 1 \cdot 0 = 0. \]

Denominator:

\[ 3(0)^2 + 2(0) = 0. \]

This is again an indeterminate form \(\frac{0}{0}\). We apply L'Hôpital's Rule again.

### Step 4: Differentiate again

The new numerator is \(-\sin x - \sec x \tan x\). The derivative is:

\[ -\cos x - (\sec x \tan^2 x + \sec^3 x). \]

The new denominator is \(3x^2 + 2x\), and its derivative is:

\[ 6x + 2. \]

### Step 5: Apply L'Hôpital's Rule again

Now we have:

\[ \lim_{x \to 0} \frac{-\cos x - (\sec x \tan^2 x + \sec^3 x)}{6x + 2}. \]

### Step 6: Substitute \(x = 0\) again

Substituting \(x = 0\):

Numerator:

\[ -\cos(0) - (\sec(0) \cdot 0^2 + \sec^3(0)) = -1 - 1 = -2. \]

Denominator:

\[ 6(0) + 2 = 2. \]

### Final Result

Thus, we have:

\[ \lim_{x \to 0} \frac{-2}{2} = -1. \]

So the final answer is:

\[ \lim_{x \to 0} \frac{\cos x - \sec x}{x^2 (x + 1)} = -1. \]