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

Answer

To solve the limit

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

we will follow these steps:

### Step 1: Substitute the limit
We start by substituting \( x = 0 \) directly into the expression.

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

This gives us the indeterminate form \( \frac{0}{0} \), so we need to apply L'Hôpital's Rule or manipulate the expression.

### Step 2: Rewrite the limit
We can rewrite the limit as:

\[ \lim_{x \to 0} \frac{e^{(\sin^3 x)} - \cos(\sin^3 x)}{x^3} = \lim_{x \to 0} \frac{e^{(\sin^3 x)} - 1 - (\cos(\sin^3 x) - 1)}{x^3} \]

### Step 3: Use Taylor series expansion
Next, we can use the Taylor series expansions for \( e^u \) and \( \cos(u) \) around \( u = 0 \):

- For \( e^u \):
\[ e^u \approx 1 + u + \frac{u^2}{2} + O(u^3) \]
- For \( \cos(u) \):
\[ \cos(u) \approx 1 - \frac{u^2}{2} + O(u^4) \]

Substituting \( u = \sin^3 x \):

\[ \sin x \approx x \quad \text{(for small } x\text{)} \]
\[ \sin^3 x \approx x^3 \]

### Step 4: Substitute back into the limit
Now substituting \( \sin^3 x \) into the expansions:

\[ e^{(\sin^3 x)} \approx 1 + \sin^3 x + \frac{(\sin^3 x)^2}{2} \approx 1 + x^3 + \frac{x^6}{2} \]
\[ \cos(\sin^3 x) \approx 1 - \frac{(\sin^3 x)^2}{2} \approx 1 - \frac{x^6}{2} \]

### Step 5: Combine the expansions
Now substituting these back into our limit:

\[ e^{(\sin^3 x)} - \cos(\sin^3 x) \approx \left(1 + x^3 + \frac{x^6}{2}\right) - \left(1 - \frac{x^6}{2}\right) = x^3 + x^6 \]

### Step 6: Substitute into the limit
Now we substitute this back into our limit:

\[ \lim_{x \to 0} \frac{x^3 + x^6}{x^3} = \lim_{x \to 0} \left(1 + x^3\right) = 1 \]

### Conclusion
Thus, the value of the limit is

\[ \boxed{1} \]