`lim_(x->0)(1+(asinb x)/(cosx))^(1/x),` where `a,b` are non zero constants is equal to :

Options

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

Answer

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

### Step 1: Identify the form of the limit
As \( x \to 0 \), \( \sin(bx) \) approaches \( bx \) and \( \cos(x) \) approaches \( 1 \). Therefore, the expression inside the limit approaches:
\[ 1 + \frac{a \sin(bx)}{\cos(x)} \approx 1 + \frac{a(bx)}{1} = 1 + abx \]
This means we are dealing with a limit of the form \( (1 + u)^{v} \) where \( u \to 0 \) and \( v \to \infty \).

### Step 2: Take the natural logarithm
To simplify the limit, we take the natural logarithm:
\[ L = \lim_{x \to 0} \frac{1}{x} \ln\left( 1 + \frac{a \sin(bx)}{\cos(x)} \right) \]
Using the property that \( \ln(1 + u) \approx u \) as \( u \to 0 \), we have:
\[ \ln\left( 1 + \frac{a \sin(bx)}{\cos(x)} \right) \approx \frac{a \sin(bx)}{\cos(x)} \]

### Step 3: Substitute and simplify
Now substituting back into our limit:
\[ L = \lim_{x \to 0} \frac{1}{x} \cdot \frac{a \sin(bx)}{\cos(x)} \]
As \( x \to 0 \), \( \cos(x) \to 1 \), so we can simplify:
\[ L = \lim_{x \to 0} \frac{a \sin(bx)}{x} \]

### Step 4: Use the limit property of sine
Using the limit property \( \lim_{x \to 0} \frac{\sin(kx)}{x} = k \):
\[ L = \lim_{x \to 0} a \cdot b = ab \]

### Step 5: Exponentiate to find the limit
Now, recalling that we took the logarithm earlier:
\[ \lim_{x \to 0} \left( 1 + \frac{a \sin(bx)}{\cos(x)} \right)^{\frac{1}{x}} = e^L = e^{ab} \]

### Final Answer
Thus, the limit is:
\[ \lim_{x \to 0} \left( 1 + \frac{a \sin(bx)}{\cos(x)} \right)^{\frac{1}{x}} = e^{ab} \]