`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

Options

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

Answer

To solve the limit problem given by

\[ \lim_{x \to 0} (1 + ax)^{\frac{b}{x}} = e^2 \]

where \(a\) and \(b\) are natural numbers such that \(a + b = 3\), we can follow these steps:

### Step 1: Identify the form of the limit
As \(x\) approaches 0, the expression \((1 + ax)\) approaches \(1\), and thus we have the indeterminate form \(1^{\infty}\).

### Step 2: Use the exponential limit property
For limits of the form \(1^{\infty}\), we can rewrite the expression using the exponential function:

\[ \lim_{x \to 0} (1 + ax)^{\frac{b}{x}} = e^{\lim_{x \to 0} \frac{b}{x} \ln(1 + ax)} \]

### Step 3: Simplify \(\ln(1 + ax)\)
Using the Taylor series expansion for \(\ln(1 + u)\) around \(u = 0\):

\[ \ln(1 + ax) \approx ax \quad \text{as } x \to 0 \]

### Step 4: Substitute back into the limit
Now substituting this back into our limit gives:

\[ \lim_{x \to 0} \frac{b}{x} \ln(1 + ax) \approx \lim_{x \to 0} \frac{b}{x} (ax) = \lim_{x \to 0} ab = ab \]

### Step 5: Set the limit equal to \(e^2\)
From the limit, we have:

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

### Step 6: Set up the system of equations
Now we have two equations:

1. \(ab = 2\)
2. \(a + b = 3\)

### Step 7: Solve for \(a\) and \(b\)
From the second equation, we can express \(a\) in terms of \(b\):

\[ a = 3 - b \]

Substituting this into the first equation:

\[ (3 - b)b = 2 \]

Expanding this gives:

\[ 3b - b^2 = 2 \]

Rearranging leads to:

\[ b^2 - 3b + 2 = 0 \]

### Step 8: Factor the quadratic
Factoring the quadratic equation:

\[ (b - 1)(b - 2) = 0 \]

This gives us the roots:

\[ b = 1 \quad \text{or} \quad b = 2 \]

### Step 9: Find corresponding values of \(a\)
Using \(b = 1\):

\[ a = 3 - 1 = 2 \]

Using \(b = 2\):

\[ a = 3 - 2 = 1 \]

### Final Result
Thus, the ordered pairs \((a, b)\) that satisfy the conditions are:

\[ (2, 1) \quad \text{and} \quad (1, 2) \]