Evaluate `lim_(xto0)(ae^(x)-b)/x=2`. Find a and b

Answer

To evaluate the limit \( \lim_{x \to 0} \frac{ae^x - b}{x} = 2 \) and find the values of \( a \) and \( b \), we can follow these steps:

### Step 1: Set up the limit
We start with the limit expression:
\[ \lim_{x \to 0} \frac{ae^x - b}{x} = 2 \]

### Step 2: Substitute \( x = 0 \)
Substituting \( x = 0 \) directly into the expression gives:
\[ \frac{ae^0 - b}{0} = \frac{a - b}{0} \]
This results in an indeterminate form \( \frac{0}{0} \), which means we need to apply L'Hôpital's Rule or ensure the numerator approaches zero.

### Step 3: Set the numerator to zero
For the limit to be of the form \( \frac{0}{0} \), we need:
\[ a - b = 0 \]
This gives us our first equation:
\[ a = b \quad \text{(Equation 1)} \]

### Step 4: Differentiate the numerator and denominator
Now we differentiate the numerator and denominator:
\[ \text{Numerator: } \frac{d}{dx}(ae^x - b) = ae^x \]
\[ \text{Denominator: } \frac{d}{dx}(x) = 1 \]
Now we can apply L'Hôpital's Rule:
\[ \lim_{x \to 0} \frac{ae^x}{1} = a \]

### Step 5: Set the limit equal to 2
From the limit we have:
\[ a = 2 \]

### Step 6: Substitute \( a \) back into Equation 1
Now we substitute \( a \) back into Equation 1:
\[ a = b \implies 2 = b \]

### Conclusion
Thus, we find:
\[ a = 2 \quad \text{and} \quad b = 2 \]

### Final Answer
The values of \( a \) and \( b \) are:
\[ a = 2, \quad b = 2 \]