The value of `lim_(xrarr0) (sinx-x+(x^3)/(6))/(x^5)`, is

Options

A. 0
B. 1
C. `1//60`
D. `1//120`

Answer

To find the limit

\[ \lim_{x \to 0} \frac{\sin x - x + \frac{x^3}{6}}{x^5}, \]

we first observe that substituting \(x = 0\) directly into the expression results in the indeterminate form \( \frac{0}{0} \). Therefore, we will apply L'Hôpital's Rule, which states that if we have an indeterminate form \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \), we can differentiate the numerator and the denominator until we reach a determinate form.

### Step 1: Apply L'Hôpital's Rule
Differentiate the numerator and the denominator:

- The numerator:
\[ \frac{d}{dx}(\sin x - x + \frac{x^3}{6}) = \cos x - 1 + \frac{1}{2} x^2. \]

- The denominator:
\[ \frac{d}{dx}(x^5) = 5x^4. \]

Thus, we rewrite the limit as:
\[ \lim_{x \to 0} \frac{\cos x - 1 + \frac{1}{2} x^2}{5x^4}. \]

### Step 2: Substitute \(x = 0\) Again
Substituting \(x = 0\) into the new expression gives:
\[ \frac{\cos(0) - 1 + \frac{1}{2} \cdot 0^2}{5 \cdot 0^4} = \frac{1 - 1 + 0}{0} = \frac{0}{0}. \]
This is still an indeterminate form, so we apply L'Hôpital's Rule again.

### Step 3: Apply L'Hôpital's Rule Again
Differentiate the numerator and the denominator again:

- The numerator:
\[ \frac{d}{dx}(\cos x - 1 + \frac{1}{2} x^2) = -\sin x + x. \]

- The denominator:
\[ \frac{d}{dx}(5x^4) = 20x^3. \]

Now, we have:
\[ \lim_{x \to 0} \frac{-\sin x + x}{20x^3}. \]

### Step 4: Substitute \(x = 0\) Again
Substituting \(x = 0\) gives:
\[ \frac{-\sin(0) + 0}{20 \cdot 0^3} = \frac{0}{0}. \]
This is still an indeterminate form, so we apply L'Hôpital's Rule once more.

### Step 5: Apply L'Hôpital's Rule Again
Differentiate the numerator and the denominator again:

- The numerator:
\[ \frac{d}{dx}(-\sin x + x) = -\cos x + 1. \]

- The denominator:
\[ \frac{d}{dx}(20x^3) = 60x^2. \]

Now, we have:
\[ \lim_{x \to 0} \frac{-\cos x + 1}{60x^2}. \]

### Step 6: Substitute \(x = 0\) Again
Substituting \(x = 0\) gives:
\[ \frac{-\cos(0) + 1}{60 \cdot 0^2} = \frac{-1 + 1}{0} = \frac{0}{0}. \]
This is still an indeterminate form, so we apply L'Hôpital's Rule one last time.

### Step 7: Apply L'Hôpital's Rule Again
Differentiate the numerator and the denominator again:

- The numerator:
\[ \frac{d}{dx}(-\cos x + 1) = \sin x. \]

- The denominator:
\[ \frac{d}{dx}(60x^2) = 120x. \]

Now, we have:
\[ \lim_{x \to 0} \frac{\sin x}{120x}. \]

### Step 8: Substitute \(x = 0\) Again
Substituting \(x = 0\) gives:
\[ \frac{\sin(0)}{120 \cdot 0} = \frac{0}{0}. \]
This is still an indeterminate form, so we apply L'Hôpital's Rule one last time.

### Step 9: Apply L'Hôpital's Rule Again
Differentiate the numerator and the denominator again:

- The numerator:
\[ \frac{d}{dx}(\sin x) = \cos x. \]

- The denominator:
\[ \frac{d}{dx}(120x) = 120. \]

Now, we have:
\[ \lim_{x \to 0} \frac{\cos x}{120}. \]

### Step 10: Substitute \(x = 0\) Finally
Substituting \(x = 0\) gives:
\[ \frac{\cos(0)}{120} = \frac{1}{120}. \]

Thus, the final result is:

\[ \lim_{x \to 0} \frac{\sin x - x + \frac{x^3}{6}}{x^5} = \frac{1}{120}. \]