The function `f(x) = (|x+2|)/(tan^(-1)(x+2))`, is continuous for x∈R x∈R−{0} x∈R−{−2} None of these

Options

A. `x in R`
B. `x in R - {0}`
C. `x in R - {-2}`
D. None of these

Answer

To determine the continuity of the function \( f(x) = \frac{|x+2|}{\tan^{-1}(x+2)} \), we need to analyze the function at the point where it may not be continuous, which is \( x = -2 \).

### Step-by-Step Solution:

1. **Identify the point of interest**:
We need to check the continuity at \( x = -2 \) since the absolute value function can create a point of discontinuity.

2. **Calculate the left-hand limit as \( x \) approaches -2**:
\[ \lim_{x \to -2^-} f(x) = \lim_{h \to 0} f(-2 - h) = \lim_{h \to 0} \frac{|(-2 - h) + 2|}{\tan^{-1}((-2 - h) + 2)} \]
Simplifying the expression:
\[ = \lim_{h \to 0} \frac{| -h |}{\tan^{-1}(-h)} = \lim_{h \to 0} \frac{h}{\tan^{-1}(-h)} \]

3. **Evaluate the limit**:
As \( h \to 0 \), \( \tan^{-1}(-h) \) approaches \( -h \). Thus:
\[ = \lim_{h \to 0} \frac{h}{-h} = -1 \]

4. **Calculate the right-hand limit as \( x \) approaches -2**:
\[ \lim_{x \to -2^+} f(x) = \lim_{h \to 0} f(-2 + h) = \lim_{h \to 0} \frac{|(-2 + h) + 2|}{\tan^{-1}((-2 + h) + 2)} \]
Simplifying the expression:
\[ = \lim_{h \to 0} \frac{|h|}{\tan^{-1}(h)} = \lim_{h \to 0} \frac{h}{\tan^{-1}(h)} \]

5. **Evaluate the limit**:
As \( h \to 0 \), \( \tan^{-1}(h) \) approaches \( h \). Thus:
\[ = \lim_{h \to 0} \frac{h}{h} = 1 \]

6. **Compare the left-hand and right-hand limits**:
We found:
- Left-hand limit: \( -1 \)
- Right-hand limit: \( 1 \)

Since the left-hand limit does not equal the right-hand limit, we conclude that the function is not continuous at \( x = -2 \).

7. **Final conclusion**:
The function \( f(x) \) is continuous for \( x \in \mathbb{R} \setminus \{-2\} \).

### Answer:
The function \( f(x) \) is continuous for \( x \in \mathbb{R} - \{-2\} \).

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