The value of `lim_(xrarr0) (e^x-(x+x))/(x^2)`,is

Options

A. 0
B. `1//2`
C. `2`
D. e

Answer

To solve the limit \( \lim_{x \to 0} \frac{e^x - (x + x)}{x^2} \), we can follow these steps:

### Step 1: Simplify the expression
First, we rewrite the expression in the limit:
\[ \lim_{x \to 0} \frac{e^x - 2x}{x^2} \]

### Step 2: Check the form of the limit
Now, we substitute \( x = 0 \) into the expression:
\[ e^0 - 2(0) = 1 - 0 = 1 \]
The denominator \( x^2 \) approaches \( 0 \) as \( x \to 0 \). Thus, we have the form \( \frac{1 - 0}{0} \), which is of the form \( \frac{1}{0} \), indicating that we should apply L'Hôpital's Rule since we have an indeterminate form.

### Step 3: Apply L'Hôpital's Rule
According to L'Hôpital's Rule, we differentiate the numerator and the denominator:
- The derivative of the numerator \( e^x - 2x \) is \( e^x - 2 \).
- The derivative of the denominator \( x^2 \) is \( 2x \).

Now we can rewrite the limit:
\[ \lim_{x \to 0} \frac{e^x - 2}{2x} \]

### Step 4: Substitute again
Now, substituting \( x = 0 \) into the new expression:
\[ \frac{e^0 - 2}{2(0)} = \frac{1 - 2}{0} = \frac{-1}{0} \]
This again indicates an indeterminate form, so we apply L'Hôpital's Rule once more.

### Step 5: Differentiate again
Differentiating again:
- The derivative of the numerator \( e^x - 2 \) is \( e^x \).
- The derivative of the denominator \( 2x \) is \( 2 \).

Now we have:
\[ \lim_{x \to 0} \frac{e^x}{2} \]

### Step 6: Substitute again
Now substituting \( x = 0 \):
\[ \frac{e^0}{2} = \frac{1}{2} \]

### Final Answer
Thus, the value of the limit is:
\[ \boxed{\frac{1}{2}} \]