Evaluate the following limits :
`lim_(x to 0)(e^(sin2x)-e^(sinx))/x`

Answer

To evaluate the limit

\[ \lim_{x \to 0} \frac{e^{\sin 2x} - e^{\sin x}}{x}, \]

we first check the form of the limit as \(x\) approaches 0.

1. **Substituting \(x = 0\)**:
\[ e^{\sin 2(0)} - e^{\sin(0)} = e^{0} - e^{0} = 1 - 1 = 0. \]
The denominator is also \(0\) since \(x = 0\). Thus, we have a \( \frac{0}{0} \) indeterminate form.

2. **Applying L'Hôpital's Rule**:
Since we have an indeterminate form, we can apply L'Hôpital's Rule, which states that if \(\lim_{x \to a} \frac{f(x)}{g(x)}\) results in \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\), then:
\[ \lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a} \frac{f'(x)}{g'(x)}, \]
provided the limit on the right exists.

Here, let:
- \(f(x) = e^{\sin 2x} - e^{\sin x}\)
- \(g(x) = x\)

We need to find the derivatives \(f'(x)\) and \(g'(x)\).

3. **Finding \(f'(x)\)**:
Using the chain rule:
\[ f'(x) = \frac{d}{dx}(e^{\sin 2x}) - \frac{d}{dx}(e^{\sin x}) = e^{\sin 2x} \cdot \cos(2x) \cdot 2 - e^{\sin x} \cdot \cos(x). \]
Thus,
\[ f'(x) = 2 e^{\sin 2x} \cos(2x) - e^{\sin x} \cos(x). \]

4. **Finding \(g'(x)\)**:
\[ g'(x) = 1. \]

5. **Applying L'Hôpital's Rule**:
Now we can substitute back into the limit:
\[ \lim_{x \to 0} \frac{f'(x)}{g'(x)} = \lim_{x \to 0} \left(2 e^{\sin 2x} \cos(2x) - e^{\sin x} \cos(x)\right). \]

6. **Substituting \(x = 0\) again**:
\[ = 2 e^{\sin(0)} \cos(0) - e^{\sin(0)} \cos(0) = 2 \cdot 1 \cdot 1 - 1 \cdot 1 = 2 - 1 = 1. \]

Thus, the final value of the limit is

\[ \boxed{1}. \]