If `lim_(x to 0) (1+ ax)^(b//x) =e^(2)`, where `a,b in N` such that `a+b=3`, then the value of (a,b) is equal to:

Options

A. (16,8)
B. (8,4)
C. (2,1)
D. (1,2)

Answer

To solve the problem, we need to evaluate the limit given and find the values of \(a\) and \(b\) such that \(a + b = 3\) and the limit equals \(e^2\).

### Step 1: Set up the limit expression
We start with the limit:
\[ \lim_{x \to 0} (1 + ax)^{\frac{b}{x}} = e^2 \]

### Step 2: Identify the form of the limit
As \(x\) approaches 0, the expression \(1 + ax\) approaches 1, and thus \((1 + ax)^{\frac{b}{x}}\) approaches the indeterminate form \(1^\infty\). We can use the limit property:
\[ \lim_{x \to 0} (1 + u)^{\frac{v}{u}} = e^{\lim_{u \to 0} v} \]
where \(u = ax\) and \(v = b\).

### Step 3: Rewrite the limit
Using the above property, we rewrite the limit:
\[ \lim_{x \to 0} (1 + ax)^{\frac{b}{x}} = e^{\lim_{x \to 0} \frac{b \cdot ax}{ax}} = e^{\lim_{x \to 0} \frac{b}{a}} = e^{\frac{b}{a}} \]
Setting this equal to \(e^2\):
\[ e^{\frac{b}{a}} = e^2 \]

### Step 4: Equate the exponents
Since the bases are the same, we equate the exponents:
\[ \frac{b}{a} = 2 \]
This implies:
\[ b = 2a \]

### Step 5: Use the condition \(a + b = 3\)
We also have the second equation from the problem:
\[ a + b = 3 \]
Substituting \(b = 2a\) into this equation:
\[ a + 2a = 3 \]
\[ 3a = 3 \]
\[ a = 1 \]

### Step 6: Find \(b\)
Now substituting \(a = 1\) back into \(b = 2a\):
\[ b = 2 \cdot 1 = 2 \]

### Final Result
Thus, the values of \(a\) and \(b\) are:
\[ (a, b) = (1, 2) \]