The value of `lim_(x to 0) (e^("sin"^(3) x) - cos (("sin" x)^(3)))/(x^(3))` is ____

Answer

To solve the limit

\[ \lim_{x \to 0} \frac{e^{(\sin^3 x)} - \cos((\sin x)^3)}{x^3}, \]

we will follow these steps:

### Step 1: Evaluate the limit directly
First, we substitute \(x = 0\) into the expression:

\[ e^{(\sin^3 0)} - \cos((\sin 0)^3) = e^0 - \cos(0) = 1 - 1 = 0. \]

Since both the numerator and denominator approach 0, we have an indeterminate form \( \frac{0}{0} \).

### Step 2: Apply L'Hôpital's Rule
Since we have the indeterminate form \( \frac{0}{0} \), we can apply L'Hôpital's Rule, which states that:

\[ \lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a} \frac{f'(x)}{g'(x)} \]

if the limit on the right exists. We need to differentiate the numerator and the denominator.

#### Differentiate the numerator:
Let \(f(x) = e^{(\sin^3 x)} - \cos((\sin x)^3)\).

Using the chain rule:
1. The derivative of \(e^{(\sin^3 x)}\) is \(e^{(\sin^3 x)} \cdot \frac{d}{dx}(\sin^3 x) = e^{(\sin^3 x)} \cdot 3\sin^2 x \cdot \cos x\).
2. The derivative of \(-\cos((\sin x)^3)\) is \(\sin((\sin x)^3) \cdot 3(\sin x)^2 \cdot \cos x\) (using the chain rule).

Thus, the derivative of the numerator is:

\[ f'(x) = e^{(\sin^3 x)} \cdot 3\sin^2 x \cdot \cos x + \sin((\sin x)^3) \cdot 3(\sin x)^2 \cdot \cos x. \]

#### Differentiate the denominator:
The derivative of \(g(x) = x^3\) is \(g'(x) = 3x^2\).

### Step 3: Rewrite the limit
Now we can rewrite the limit using L'Hôpital's Rule:

\[ \lim_{x \to 0} \frac{f'(x)}{g'(x)} = \lim_{x \to 0} \frac{e^{(\sin^3 x)} \cdot 3\sin^2 x \cdot \cos x + \sin((\sin x)^3) \cdot 3(\sin x)^2 \cdot \cos x}{3x^2}. \]

### Step 4: Simplify the limit
We can factor out \(3\) from the numerator:

\[ = \lim_{x \to 0} \frac{3\left(e^{(\sin^3 x)} \sin^2 x \cos x + \sin((\sin x)^3) (\sin x)^2 \cos x\right)}{3x^2}. \]

Canceling the \(3\) gives:

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

### Step 5: Evaluate the limit
Now we can evaluate the limit as \(x\) approaches \(0\):

1. As \(x \to 0\), \(e^{(\sin^3 x)} \to e^0 = 1\).
2. \(\sin^2 x \to 0\).
3. \(\cos x \to 1\).

Thus, the first term \(e^{(\sin^3 x)} \sin^2 x \cos x \to 1 \cdot 0 \cdot 1 = 0\).

For the second term, \(\sin((\sin x)^3) \to \sin(0) = 0\) and \((\sin x)^2 \to 0\), so this term also approaches \(0\).

Therefore, the limit simplifies to:

\[ \lim_{x \to 0} \frac{0 + 0}{0} = 0. \]

### Final Result
Thus, the value of the limit is:

\[ \boxed{1}. \]