The value of `lim_(xrarr 0) (e^x+log (1+x)-(1-x)^-2)/(x^2)` is equal to

Options

A. 0
B. -3
C. -1
D. infinity

Explanation

To find the limit

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

we start by substituting \( x = 0 \) into the expression.

### Step 1: Evaluate the limit directly
Substituting \( x = 0 \):

- \( e^0 = 1 \)
- \( \log(1+0) = \log(1) = 0 \)
- \( (1-0)^{-2} = 1^{-2} = 1 \)

Thus, the numerator becomes:

\[ 1 + 0 - 1 = 0. \]

The denominator is:

\[ 0^2 = 0. \]

This gives us the indeterminate form \( \frac{0}{0} \).

### Step 2: Apply L'Hôpital's Rule
Since we have the form \( \frac{0}{0} \), we can apply L'Hôpital's Rule, which states that we can take the derivative of the numerator and the denominator.

The limit now becomes:

\[ \lim_{x \to 0} \frac{d}{dx}(e^x + \log(1+x) - (1-x)^{-2}) \bigg/ \frac{d}{dx}(x^2). \]

### Step 3: Differentiate the numerator and denominator
1. Differentiate the numerator:
- The derivative of \( e^x \) is \( e^x \).
- The derivative of \( \log(1+x) \) is \( \frac{1}{1+x} \).
- The derivative of \( (1-x)^{-2} \) using the chain rule is:
\[ -2(1-x)^{-3} \cdot (-1) = \frac{2}{(1-x)^3}. \]

So, the derivative of the numerator is:

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

2. Differentiate the denominator:
- The derivative of \( x^2 \) is \( 2x \).

### Step 4: Rewrite the limit
Now, we rewrite the limit:

\[ \lim_{x \to 0} \frac{e^x + \frac{1}{1+x} + \frac{2}{(1-x)^3}}{2x}. \]

### Step 5: Substitute \( x = 0 \) again
Substituting \( x = 0 \):

- \( e^0 = 1 \)
- \( \frac{1}{1+0} = 1 \)
- \( \frac{2}{(1-0)^3} = 2 \)

Thus, the numerator becomes:

\[ 1 + 1 + 2 = 4. \]

The denominator is:

\[ 2 \cdot 0 = 0. \]

This still gives us the form \( \frac{4}{0} \), which indicates we need to apply L'Hôpital's Rule again.

### Step 6: Apply L'Hôpital's Rule again
Differentiate the numerator and denominator again:

1. Differentiate the numerator:
- The derivative of \( e^x \) is \( e^x \).
- The derivative of \( \frac{1}{1+x} \) is \( -\frac{1}{(1+x)^2} \).
- The derivative of \( \frac{2}{(1-x)^3} \) is \( -\frac{6}{(1-x)^4} \).

So, the new derivative of the numerator is:

\[ e^x - \frac{1}{(1+x)^2} - \frac{6}{(1-x)^4}. \]

2. Differentiate the denominator:
- The derivative of \( 2x \) is \( 2 \).

### Step 7: Rewrite the limit again
Now we have:

\[ \lim_{x \to 0} \frac{e^x - \frac{1}{(1+x)^2} - \frac{6}{(1-x)^4}}{2}. \]

### Step 8: Substitute \( x = 0 \) again
Substituting \( x = 0 \):

- \( e^0 = 1 \)
- \( -\frac{1}{(1+0)^2} = -1 \)
- \( -\frac{6}{(1-0)^4} = -6 \)

Thus, the numerator becomes:

\[ 1 - 1 - 6 = -6. \]

The denominator is:

\[ 2. \]

### Final Step: Calculate the limit
Now we can calculate the limit:

\[ \lim_{x \to 0} \frac{-6}{2} = -3. \]

Thus, the value of the limit is:

\[ \boxed{-3}. \]