Value of `underset(x to 0)lim (a^(sin x)-1)/(sin x)` is

Options

A. log a
B. sin x
C. log(sin x)
D. cos x

Answer

To find the value of the limit

\[ \lim_{x \to 0} \frac{a^{\sin x} - 1}{\sin x}, \]

we start by substituting \( x = 0 \):

1. **Substitution**:
\[ a^{\sin(0)} - 1 = a^0 - 1 = 1 - 1 = 0, \]
and
\[ \sin(0) = 0. \]
This gives us the indeterminate form \( \frac{0}{0} \).

**Hint**: Check if substituting \( x = 0 \) leads to an indeterminate form.

2. **Applying L'Hôpital's Rule**: Since we have the \( \frac{0}{0} \) form, we can apply L'Hôpital's Rule, which states that we can take the derivative of the numerator and the derivative of the denominator.

- **Numerator**: The derivative of \( a^{\sin x} - 1 \) with respect to \( x \) is:
\[ \frac{d}{dx}(a^{\sin x}) = a^{\sin x} \cdot \ln(a) \cdot \cos x. \]
- **Denominator**: The derivative of \( \sin x \) is:
\[ \frac{d}{dx}(\sin x) = \cos x. \]

3. **Rewriting the limit**: Now we can rewrite the limit using L'Hôpital's Rule:
\[ \lim_{x \to 0} \frac{a^{\sin x} \cdot \ln(a) \cdot \cos x}{\cos x}. \]

4. **Simplifying**: The \( \cos x \) terms in the numerator and denominator cancel out:
\[ \lim_{x \to 0} a^{\sin x} \cdot \ln(a). \]

5. **Substituting \( x = 0 \)** again:
\[ a^{\sin(0)} \cdot \ln(a) = a^0 \cdot \ln(a) = 1 \cdot \ln(a) = \ln(a). \]

Thus, the final value of the limit is:

\[ \boxed{\ln(a)}. \]

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