If `llim_(x to 0) (cos x + a sin bx)^(1//x)= e^(2)`, then (a,b) is equal to:

Options

A. (1,2)
B. `(2,1/2)`
C. `(2sqrt(3), 1/sqrt(3))`
D. (4,2)

Answer

To solve the limit problem, we start with the expression given in the question:

\[ \lim_{x \to 0} ( \cos x + a \sin(bx) )^{\frac{1}{x}} = e^2 \]

### Step 1: Evaluate the limit of the inner function as \( x \to 0 \)

As \( x \) approaches 0, we can use the Taylor series expansions for \( \cos x \) and \( \sin(bx) \):

\[ \cos x \approx 1 - \frac{x^2}{2} + O(x^4) \]
\[ \sin(bx) \approx bx + O(x^3) \]

Substituting these approximations into the limit gives:

\[ \cos x + a \sin(bx) \approx \left(1 - \frac{x^2}{2}\right) + a(bx) = 1 + abx - \frac{x^2}{2} + O(x^3) \]

### Step 2: Simplify the expression

As \( x \to 0 \), the dominant term in the expression is \( 1 + abx \). Therefore, we can rewrite the limit:

\[ \lim_{x \to 0} (1 + abx)^{\frac{1}{x}} \]

### Step 3: Use the exponential limit property

We know that:

\[ \lim_{x \to 0} (1 + kx)^{\frac{1}{x}} = e^k \]

Applying this property, we have:

\[ \lim_{x \to 0} (1 + abx)^{\frac{1}{x}} = e^{ab} \]

### Step 4: Set the limit equal to \( e^2 \)

From the problem statement, we have:

\[ e^{ab} = e^2 \]

### Step 5: Solve for \( ab \)

Taking the natural logarithm of both sides gives:

\[ ab = 2 \]

### Step 6: Determine values for \( a \) and \( b \)

At this point, we have one equation \( ab = 2 \). To find specific values for \( a \) and \( b \), we can choose \( a = 2 \) and \( b = 1 \) (or any other pairs such as \( a = 1 \) and \( b = 2 \)).

Thus, one possible solution is:

\[ (a, b) = (2, 1) \]

### Final Answer

\[ (a, b) = (2, 1) \]

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