The value of `lim_(x->0)(1-1/2^x)(1/(sqrt(tanx+4)-2))`

Options

A. ` log_(a) 16`
B. Cannot exist
C. 3l n2
D. 4 ln 2

Answer

To solve the limit \( \lim_{x \to 0} \left(1 - \frac{1}{2^x}\right) \left(\frac{1}{\sqrt{\tan x + 4} - 2}\right) \), we will follow these steps:

### Step 1: Rewrite the limit
We start with the limit:
\[ \lim_{x \to 0} \left(1 - \frac{1}{2^x}\right) \left(\frac{1}{\sqrt{\tan x + 4} - 2}\right) \]

### Step 2: Evaluate the limit directly
As \( x \to 0 \):
- \( 2^x \to 1 \) so \( 1 - \frac{1}{2^x} \to 0 \)
- \( \tan x \to 0 \) so \( \sqrt{\tan x + 4} \to \sqrt{4} = 2 \) thus \( \sqrt{\tan x + 4} - 2 \to 0 \)

This gives us the indeterminate form \( \frac{0}{0} \).

### Step 3: Apply L'Hôpital's Rule
Since we have an indeterminate form, we can apply L'Hôpital's Rule. We need to differentiate the numerator and the denominator.

#### Numerator:
The numerator is \( 1 - \frac{1}{2^x} \). The derivative is:
\[ \frac{d}{dx}\left(1 - \frac{1}{2^x}\right) = 0 + \frac{\ln(2)}{2^x} \]

#### Denominator:
The denominator is \( \sqrt{\tan x + 4} - 2 \). The derivative is:
\[ \frac{d}{dx}\left(\sqrt{\tan x + 4} - 2\right) = \frac{1}{2\sqrt{\tan x + 4}} \cdot \sec^2 x \]

### Step 4: Rewrite the limit using derivatives
Now we rewrite the limit using these derivatives:
\[ \lim_{x \to 0} \frac{\frac{\ln(2)}{2^x}}{\frac{1}{2\sqrt{\tan x + 4}} \cdot \sec^2 x} \]

### Step 5: Evaluate the limit again
As \( x \to 0 \):
- \( 2^x \to 1 \) so \( \frac{\ln(2)}{2^x} \to \ln(2) \)
- \( \tan x \to 0 \) so \( \sqrt{\tan x + 4} \to 2 \) and \( \sec^2 x \to 1 \)

Thus, substituting these values gives:
\[ \lim_{x \to 0} \frac{\ln(2)}{\frac{1}{2 \cdot 2} \cdot 1} = \frac{\ln(2)}{\frac{1}{4}} = 4 \ln(2) \]

### Final Answer
Therefore, the value of the limit is:
\[ \lim_{x \to 0} \left(1 - \frac{1}{2^x}\right) \left(\frac{1}{\sqrt{\tan x + 4} - 2}\right) = 4 \ln(2) \]