If `f(x)=(x+1)^(cotx)` is continuous at x = 0, then what is f (0) equal to?

Options

A. 1
B. e
C. `(1)/(e)`
D. `e^(2)`

Answer

To find \( f(0) \) for the function \( f(x) = (x + 1)^{\cot x} \) and ensure it is continuous at \( x = 0 \), we will follow these steps:

### Step 1: Understand the Continuity Condition
A function is continuous at a point if the limit of the function as it approaches that point is equal to the function's value at that point. Therefore, we need to find:
\[ \lim_{x \to 0} f(x) = f(0) \]

### Step 2: Substitute the Function
We substitute \( f(x) \) into the limit:
\[ \lim_{x \to 0} f(x) = \lim_{x \to 0} (x + 1)^{\cot x} \]

### Step 3: Rewrite the Expression
To evaluate this limit, we can rewrite it using the exponential function:
\[ f(x) = (x + 1)^{\cot x} = e^{\cot x \cdot \ln(x + 1)} \]
Thus, we need to evaluate:
\[ \lim_{x \to 0} e^{\cot x \cdot \ln(x + 1)} \]

### Step 4: Evaluate the Limit Inside the Exponential
We need to find:
\[ \lim_{x \to 0} \cot x \cdot \ln(x + 1) \]
As \( x \to 0 \), \( \cot x \) approaches \( \frac{1}{x} \) and \( \ln(x + 1) \) approaches \( x \):
\[ \cot x \approx \frac{1}{x} \quad \text{and} \quad \ln(x + 1) \approx x \]
Thus, we have:
\[ \cot x \cdot \ln(x + 1) \approx \frac{1}{x} \cdot x = 1 \]

### Step 5: Final Limit Calculation
Now we can compute the limit:
\[ \lim_{x \to 0} \cot x \cdot \ln(x + 1) = 1 \]
Therefore:
\[ \lim_{x \to 0} e^{\cot x \cdot \ln(x + 1)} = e^1 = e \]

### Step 6: Conclusion
Since \( \lim_{x \to 0} f(x) = e \), we conclude that:
\[ f(0) = e \]

### Final Answer
Thus, \( f(0) = e \).

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