If `lim_(xrarr0)(axe^(x)-b log(1+x))/(x^(2))=3`, then the value of a and b are respectively.

Options

A. 2, 2
B. 1, 2
C. 2, 1
D. 2, 0

Answer

To solve the limit problem given by

\[ \lim_{x \to 0} \frac{a x e^x - b \log(1+x)}{x^2} = 3, \]

we will follow these steps:

### Step 1: Expand \(e^x\) and \(\log(1+x)\)

Using Taylor series expansions around \(x = 0\):

- The expansion of \(e^x\) is:

\[ e^x = 1 + x + \frac{x^2}{2} + O(x^3). \]

- The expansion of \(\log(1+x)\) is:

\[ \log(1+x) = x - \frac{x^2}{2} + O(x^3). \]

### Step 2: Substitute the expansions into the limit

Substituting these expansions into the expression we have:

\[ a x e^x = a x \left(1 + x + \frac{x^2}{2} + O(x^3)\right) = a x + a x^2 + \frac{a x^3}{2} + O(x^4), \]

\[ b \log(1+x) = b \left(x - \frac{x^2}{2} + O(x^3)\right) = b x - \frac{b x^2}{2} + O(x^3). \]

### Step 3: Combine the terms

Now, we can combine these results in the limit:

\[ a x e^x - b \log(1+x) = (a x + a x^2 + \frac{a x^3}{2}) - (b x - \frac{b x^2}{2}) + O(x^3). \]

This simplifies to:

\[ (a - b)x + \left(a + \frac{b}{2}\right)x^2 + O(x^3). \]

### Step 4: Write the limit expression

Now, we can write the limit expression:

\[ \frac{(a - b)x + \left(a + \frac{b}{2}\right)x^2 + O(x^3)}{x^2}. \]

This simplifies to:

\[ \frac{(a - b)}{x} + \left(a + \frac{b}{2}\right) + O(x). \]

### Step 5: Evaluate the limit as \(x \to 0\)

For the limit to exist and equal 3, the coefficient of \(\frac{1}{x}\) must be zero, which gives us the first equation:

\[ a - b = 0 \quad \Rightarrow \quad a = b. \]

The remaining term must equal 3:

\[ a + \frac{b}{2} = 3. \]

### Step 6: Substitute \(b = a\) into the second equation

Substituting \(b = a\) into the second equation:

\[ a + \frac{a}{2} = 3 \quad \Rightarrow \quad \frac{3a}{2} = 3 \quad \Rightarrow \quad 3a = 6 \quad \Rightarrow \quad a = 2. \]

### Step 7: Find \(b\)

Since \(a = b\), we have:

\[ b = 2. \]

### Final Result

Thus, the values of \(a\) and \(b\) are:

\[ \boxed{(2, 2)}. \]