`lim_(x to 2) (2^(x) + 2^(2 -x) - 5)/((1)/(sqrt(2^(x))) + lambda (2)^(1 - x)) (lambda in R)` has non zero value, which can be

Explanation

To solve the limit problem:

\[ \lim_{x \to 2} \frac{2^x + 2^{2-x} - 5}{\frac{1}{\sqrt{2^x}} + \lambda \cdot 2^{1-x}} \]

where \(\lambda \in \mathbb{R}\) has a non-zero value, we will follow these steps:

### Step 1: Substitute \(x = 2\)

First, we substitute \(x = 2\) into the expression to check if we get an indeterminate form.

**Numerator:**
\[ 2^2 + 2^{2-2} - 5 = 4 + 1 - 5 = 0 \]

**Denominator:**
\[ \frac{1}{\sqrt{2^2}} + \lambda \cdot 2^{1-2} = \frac{1}{\sqrt{4}} + \lambda \cdot 2^{-1} = \frac{1}{2} + \frac{\lambda}{2} \]

Both the numerator and denominator evaluate to 0, giving us the indeterminate form \( \frac{0}{0} \).

### Step 2: Apply L'Hôpital's Rule

Since we have the \( \frac{0}{0} \) form, we can apply L'Hôpital's Rule, which states that we can take the derivative of the numerator and the derivative of the denominator separately.

**Differentiate the Numerator:**
\[ \frac{d}{dx}(2^x + 2^{2-x} - 5) = 2^x \ln(2) + 2^{2-x} \cdot (-\ln(2)) = 2^x \ln(2) - 2^{2-x} \ln(2) \]

**Differentiate the Denominator:**
\[ \frac{d}{dx}\left(\frac{1}{\sqrt{2^x}} + \lambda \cdot 2^{1-x}\right) = -\frac{1}{2} \cdot 2^{-x/2} \cdot \ln(2) + \lambda \cdot (-\ln(2) \cdot 2^{1-x}) \]

### Step 3: Substitute \(x = 2\) Again

Now we substitute \(x = 2\) into the derivatives.

**Numerator:**
\[ 2^2 \ln(2) - 2^{2-2} \ln(2) = 4 \ln(2) - 1 \ln(2) = 3 \ln(2) \]

**Denominator:**
\[ -\frac{1}{2} \cdot 2^{-2/2} \cdot \ln(2) - \lambda \cdot \ln(2) \cdot 2^{1-2} = -\frac{1}{2} \cdot \frac{1}{2} \ln(2) - \lambda \cdot \ln(2) \cdot \frac{1}{2} = -\frac{1}{4} \ln(2) - \frac{\lambda}{2} \ln(2) \]

### Step 4: Set the Denominator to Not Equal Zero

For the limit to be non-zero, the denominator must not be zero:

\[ -\frac{1}{4} \ln(2) - \frac{\lambda}{2} \ln(2) \neq 0 \]

This implies:

\[ -\frac{1}{4} - \frac{\lambda}{2} \neq 0 \implies \lambda \neq -\frac{1}{2} \]

### Step 5: Final Limit Calculation

Now we can write the limit:

\[ \lim_{x \to 2} \frac{3 \ln(2)}{-\frac{1}{4} \ln(2) - \frac{\lambda}{2} \ln(2)} = \frac{3 \ln(2)}{-\frac{1}{4} \ln(2) - \frac{\lambda}{2} \ln(2)} \]

This simplifies to:

\[ \frac{3}{-\frac{1}{4} - \frac{\lambda}{2}} \]

### Step 6: Solve for \(\lambda\)

To find a non-zero value for \(\lambda\), we can set \(\lambda = -1\):

\[ \frac{3}{-\frac{1}{4} + \frac{1}{2}} = \frac{3}{-\frac{1}{4} + \frac{2}{4}} = \frac{3}{\frac{1}{4}} = 12 \]

Thus, the value of \(\lambda\) that gives a non-zero limit is:

\[ \lambda = -1 \]