The value of `lim_(xrarr0)(secx+tanx)^(1/x)` is equal to

Options

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

Answer

To find the value of the limit \( \lim_{x \to 0} ( \sec x + \tan x )^{\frac{1}{x}} \), we can follow these steps:

### Step 1: Rewrite the expression
We start with the expression inside the limit:
\[ \sec x + \tan x = \frac{1}{\cos x} + \frac{\sin x}{\cos x} = \frac{1 + \sin x}{\cos x} \]
Thus, we can rewrite the limit as:
\[ \lim_{x \to 0} \left( \frac{1 + \sin x}{\cos x} \right)^{\frac{1}{x}} \]

### Step 2: Simplify the limit
Now we can express the limit as:
\[ \lim_{x \to 0} \left( \frac{1 + \sin x}{\cos x} \right)^{\frac{1}{x}} = \lim_{x \to 0} \left( 1 + \sin x \right)^{\frac{1}{x}} \cdot \left( \cos x \right)^{-\frac{1}{x}} \]

### Step 3: Analyze each part
We know that as \( x \to 0 \):
- \( \sin x \to 0 \)
- \( \cos x \to 1 \)

Thus, we can analyze:
1. \( \left( 1 + \sin x \right)^{\frac{1}{x}} \)
2. \( \left( \cos x \right)^{-\frac{1}{x}} \)

### Step 4: Use the limit properties
The first part \( \left( 1 + \sin x \right)^{\frac{1}{x}} \) can be rewritten using the fact that \( \sin x \approx x \) as \( x \to 0 \):
\[ \lim_{x \to 0} \left( 1 + \sin x \right)^{\frac{1}{x}} = \lim_{x \to 0} \left( 1 + x \right)^{\frac{1}{x}} = e \]

The second part \( \left( \cos x \right)^{-\frac{1}{x}} \) approaches:
\[ \lim_{x \to 0} \left( \cos x \right)^{-\frac{1}{x}} = \lim_{x \to 0} \left( 1 \right)^{-\frac{1}{x}} = 1 \]

### Step 5: Combine the results
Combining both parts, we have:
\[ \lim_{x \to 0} ( \sec x + \tan x )^{\frac{1}{x}} = e \cdot 1 = e \]

### Final Answer
Thus, the value of the limit is:
\[ \boxed{e} \]