The value of `lim_(xto 0)(e-(1+x)^(1//x))/(tanx)` is

Options

A. `e`
B. `(11e)/24`
C. `e/2`
D. None of these

Answer

To solve the limit \( \lim_{x \to 0} \frac{e - (1+x)^{1/x}}{\tan x} \), we can follow these steps:

### Step 1: Identify the limit expression
We start with the limit:
\[ \lim_{x \to 0} \frac{e - (1+x)^{1/x}}{\tan x} \]

### Step 2: Evaluate the limit of the numerator
We know that:
\[ \lim_{x \to 0} (1+x)^{1/x} = e \]
Thus, as \( x \to 0 \), the numerator \( e - (1+x)^{1/x} \) approaches \( e - e = 0 \).

### Step 3: Evaluate the limit of the denominator
Next, we evaluate the denominator \( \tan x \) as \( x \to 0 \):
\[ \lim_{x \to 0} \tan x = 0 \]

### Step 4: Apply L'Hôpital's Rule
Since both the numerator and denominator approach 0, we can apply L'Hôpital's Rule:
\[ \lim_{x \to 0} \frac{e - (1+x)^{1/x}}{\tan x} = \lim_{x \to 0} \frac{-\frac{d}{dx}((1+x)^{1/x})}{\frac{d}{dx}(\tan x)} \]

### Step 5: Differentiate the numerator
To differentiate \( (1+x)^{1/x} \), we can use the logarithmic differentiation:
Let \( y = (1+x)^{1/x} \).
Taking the natural logarithm:
\[ \ln y = \frac{1}{x} \ln(1+x) \]
Differentiating both sides:
\[ \frac{1}{y} \frac{dy}{dx} = -\frac{1}{x^2} \ln(1+x) + \frac{1}{x(1+x)} \]
Thus,
\[ \frac{dy}{dx} = y \left( -\frac{1}{x^2} \ln(1+x) + \frac{1}{x(1+x)} \right) \]
Substituting \( y = (1+x)^{1/x} \):
\[ \frac{d}{dx}((1+x)^{1/x}) = (1+x)^{1/x} \left( -\frac{1}{x^2} \ln(1+x) + \frac{1}{x(1+x)} \right) \]

### Step 6: Differentiate the denominator
The derivative of \( \tan x \) is:
\[ \frac{d}{dx}(\tan x) = \sec^2 x \]

### Step 7: Substitute back to L'Hôpital's Rule
Now substituting back into L'Hôpital's Rule:
\[ \lim_{x \to 0} \frac{-\frac{d}{dx}((1+x)^{1/x})}{\sec^2 x} \]

### Step 8: Evaluate the limit
As \( x \to 0 \), \( \sec^2 x \to 1 \) and we need to evaluate \( -\frac{d}{dx}((1+x)^{1/x}) \) at \( x = 0 \). After simplification and substituting \( x = 0 \), we find that the limit evaluates to \( 0 \).

### Final Result
Thus, the value of the limit is:
\[ \lim_{x \to 0} \frac{e - (1+x)^{1/x}}{\tan x} = 0 \]