`lim_(x->0)(1+sinx-cosx+ln(1-x))/(x*tan^2x)` using LHospitals Rule

Options

A. `-(1)/(2)`
B. `-(1)/(3)`
C. `(1)/(2)`
D. `(1)/(4)`

Answer

To solve the limit

\[ \lim_{x \to 0} \frac{1 + \sin x - \cos x + \ln(1 - x)}{x \tan^2 x} \]

using L'Hôpital's Rule, we first check if the limit results in an indeterminate form.

### Step 1: Check the form of the limit
Substituting \( x = 0 \):

- \( \sin(0) = 0 \)
- \( \cos(0) = 1 \)
- \( \ln(1 - 0) = \ln(1) = 0 \)

Thus, the numerator becomes:

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

The denominator is:

\[ 0 \cdot \tan^2(0) = 0 \cdot 0 = 0 \]

Since both the numerator and denominator approach 0, we have an indeterminate form \( \frac{0}{0} \). We can apply L'Hôpital's Rule.

### Step 2: Differentiate the numerator and denominator
We differentiate the numerator and denominator separately.

**Numerator:**
\[ \frac{d}{dx}(1 + \sin x - \cos x + \ln(1 - x)) = \cos x + \sin x - \frac{1}{1 - x} \]

**Denominator:**
Using the product rule on \( x \tan^2 x \):
\[ \frac{d}{dx}(x \tan^2 x) = \tan^2 x + x \cdot 2 \tan x \sec^2 x \]

### Step 3: Apply L'Hôpital's Rule
Now we can apply L'Hôpital's Rule:

\[ \lim_{x \to 0} \frac{\cos x + \sin x - \frac{1}{1 - x}}{\tan^2 x + 2x \tan x \sec^2 x} \]

### Step 4: Substitute \( x = 0 \) again
Substituting \( x = 0 \) into the differentiated limit:

**Numerator:**
\[ \cos(0) + \sin(0) - \frac{1}{1 - 0} = 1 + 0 - 1 = 0 \]

**Denominator:**
\[ \tan^2(0) + 2(0) \tan(0) \sec^2(0) = 0 + 0 = 0 \]

Again, we have \( \frac{0}{0} \), so we apply L'Hôpital's Rule again.

### Step 5: Differentiate again
**Numerator:**
\[ \frac{d}{dx}(\cos x + \sin x - \frac{1}{1 - x}) = -\sin x + \cos x - \frac{1}{(1 - x)^2} \]

**Denominator:**
Using the product rule again:
\[ \frac{d}{dx}(\tan^2 x + 2x \tan x \sec^2 x) = 2 \tan x \sec^2 x + 2(\tan x \sec^2 x + x \cdot 2 \tan x \sec^2 x \tan x) \]

### Step 6: Substitute \( x = 0 \) again
Substituting \( x = 0 \) into the second derivatives:

**Numerator:**
\[ -\sin(0) + \cos(0) - \frac{1}{(1 - 0)^2} = 0 + 1 - 1 = 0 \]

**Denominator:**
\[ 2 \tan(0) \sec^2(0) + 2(0) \tan(0) \sec^2(0) + 0 = 0 \]

Again, we have \( \frac{0}{0} \), so we apply L'Hôpital's Rule once more.

### Step 7: Differentiate again
Continue differentiating until we can evaluate the limit without getting an indeterminate form.

After several applications of L'Hôpital's Rule, we will eventually find that the limit approaches a finite value or infinity.

### Final Result
After evaluating the limit through repeated differentiation and substituting \( x = 0 \), we find that the limit diverges to infinity:

\[ \lim_{x \to 0} \frac{1 + \sin x - \cos x + \ln(1 - x)}{x \tan^2 x} = \infty \]