If `l_1=lim_(xrarr-2)(x+|x|),l_2=lim_(xrarr-2)(2x+|x|)` and `l_(3)=lim_(xrarr pi//2)(cosx)/(x-pi//2)`, then

Options

A. `l_1lt l_2lt l_3`
B. `l_2lt l_3 lt l_1`
C. `l_3gt l_2 gt l_1`
D. `l_1lt l_3 lt_2`

Answer

To solve the problem step by step, we will evaluate the limits \( L_1 \), \( L_2 \), and \( L_3 \) as given in the question.

### Step 1: Evaluate \( L_1 = \lim_{x \to -2} (x + |x|) \)

1. **Identify the expression**:
\[ L_1 = \lim_{x \to -2} (x + |x|) \]

2. **Determine the value of \( |x| \)**:
Since \( x \) is approaching \(-2\) (which is negative), we have:
\[ |x| = -x \]

3. **Substitute \( |x| \) into the expression**:
\[ L_1 = \lim_{x \to -2} (x - x) = \lim_{x \to -2} 0 = 0 \]

### Step 2: Evaluate \( L_2 = \lim_{x \to -2} (2x + |x|) \)

1. **Identify the expression**:
\[ L_2 = \lim_{x \to -2} (2x + |x|) \]

2. **Determine the value of \( |x| \)**:
Again, since \( x \) is negative:
\[ |x| = -x \]

3. **Substitute \( |x| \) into the expression**:
\[ L_2 = \lim_{x \to -2} (2x - x) = \lim_{x \to -2} x \]

4. **Evaluate the limit**:
\[ L_2 = -2 \]

### Step 3: Evaluate \( L_3 = \lim_{x \to \frac{\pi}{2}} \frac{\cos x}{x - \frac{\pi}{2}} \)

1. **Identify the expression**:
\[ L_3 = \lim_{x \to \frac{\pi}{2}} \frac{\cos x}{x - \frac{\pi}{2}} \]

2. **Check the form of the limit**:
- As \( x \to \frac{\pi}{2} \), \( \cos x \to 0 \) and \( x - \frac{\pi}{2} \to 0 \). This is a \( \frac{0}{0} \) indeterminate form.

3. **Apply L'Hôpital's Rule**:
Differentiate the numerator and denominator:
- Derivative of \( \cos x \) is \( -\sin x \).
- Derivative of \( x - \frac{\pi}{2} \) is \( 1 \).

4. **Rewrite the limit using L'Hôpital's Rule**:
\[ L_3 = \lim_{x \to \frac{\pi}{2}} \frac{-\sin x}{1} \]

5. **Evaluate the limit**:
\[ L_3 = -\sin\left(\frac{\pi}{2}\right) = -1 \]

### Summary of Results
- \( L_1 = 0 \)
- \( L_2 = -2 \)
- \( L_3 = -1 \)

### Step 4: Compare the values of \( L_1, L_2, L_3 \)

- \( L_2 = -2 \)
- \( L_3 = -1 \)
- \( L_1 = 0 \)

Thus, we can conclude:
\[ L_2 < L_3 < L_1 \]