The value of `lim_(x to 0) [cos x + a sin bx]^(1//x)` is:

Options

A. 1
B. ab
C. `e^(ab)`
D. `e^(b//a)`

Answer

To solve the limit \( \lim_{x \to 0} [\cos x + a \sin(bx)]^{\frac{1}{x}} \), we can follow these steps:

### Step 1: Identify the form of the limit
First, we evaluate the expression inside the limit as \( x \) approaches 0:
\[ \cos(0) + a \sin(0) = 1 + 0 = 1. \]
Thus, we have:
\[ \lim_{x \to 0} [\cos x + a \sin(bx)]^{\frac{1}{x}} = 1^{\frac{1}{0}}. \]
This is an indeterminate form of type \( 1^{\infty} \).

### Step 2: Use the exponential limit property
To resolve the indeterminate form, we can use the property:
\[ \lim_{x \to a} f(x)^{g(x)} = e^{\lim_{x \to a} g(x) \cdot (f(x) - 1)}, \]
where \( f(x) = \cos x + a \sin(bx) \) and \( g(x) = \frac{1}{x} \).

### Step 3: Set up the limit
We rewrite the limit as:
\[ \lim_{x \to 0} [\cos x + a \sin(bx)]^{\frac{1}{x}} = e^{\lim_{x \to 0} \frac{1}{x} \left( \cos x + a \sin(bx) - 1 \right)}. \]

### Step 4: Simplify the expression inside the limit
Now, we need to evaluate:
\[ \lim_{x \to 0} \frac{\cos x + a \sin(bx) - 1}{x}. \]
Using Taylor series expansions around \( x = 0 \):
- \( \cos x \approx 1 - \frac{x^2}{2} \)
- \( \sin(bx) \approx bx \)

Thus,
\[ \cos x + a \sin(bx) \approx \left(1 - \frac{x^2}{2}\right) + a(bx) = 1 + abx - \frac{x^2}{2}. \]
So,
\[ \cos x + a \sin(bx) - 1 \approx abx - \frac{x^2}{2}. \]

### Step 5: Substitute back into the limit
Now, we substitute back:
\[ \lim_{x \to 0} \frac{abx - \frac{x^2}{2}}{x} = \lim_{x \to 0} \left(ab - \frac{x}{2}\right). \]
As \( x \to 0 \), this limit approaches \( ab \).

### Step 6: Final limit evaluation
Thus, we have:
\[ \lim_{x \to 0} [\cos x + a \sin(bx)]^{\frac{1}{x}} = e^{ab}. \]

### Conclusion
The final result is:
\[ \lim_{x \to 0} [\cos x + a \sin(bx)]^{\frac{1}{x}} = e^{ab}. \]