`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 given by

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

where \(\lambda \in \mathbb{R}\) and the limit 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:

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

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

Thus, the limit becomes:

\[ \frac{0}{\frac{1 + \lambda}{2}} \]

This is a \(0/0\) form if \(1 + \lambda = 0\) or \(\lambda = -1\).

### Step 2: Apply L'Hôpital's Rule

Since we have a \(0/0\) form when \(\lambda = -1\), we can apply L'Hôpital's Rule, which states that we can differentiate the numerator and the denominator separately.

#### Differentiate the Numerator:

The numerator is \(2^x + 2^{2-x} - 5\).

Differentiating gives:

\[ \frac{d}{dx}(2^x) = 2^x \ln(2) \]
\[ \frac{d}{dx}(2^{2-x}) = 2^{2-x} \ln(2) \cdot (-1) = -2^{2-x} \ln(2) \]

Thus, the derivative of the numerator is:

\[ 2^x \ln(2) - 2^{2-x} \ln(2) \]

#### Differentiate the Denominator:

The denominator is \(\frac{1}{\sqrt{2^x}} + \lambda \cdot 2^{1-x}\).

Differentiating gives:

\[ \frac{d}{dx}\left(\frac{1}{\sqrt{2^x}}\right) = -\frac{1}{2} \cdot 2^{-x/2} \cdot \ln(2) \]
\[ \frac{d}{dx}(\lambda \cdot 2^{1-x}) = -\lambda \cdot 2^{1-x} \ln(2) \]

Thus, the derivative of the denominator is:

\[ -\frac{1}{2} \cdot 2^{-x/2} \cdot \ln(2) - \lambda \cdot 2^{1-x} \ln(2) \]

### Step 3: Rewrite the Limit

Now we can rewrite the limit as:

\[ \lim_{x \to 2} \frac{2^x \ln(2) - 2^{2-x} \ln(2)}{-\frac{1}{2} \cdot 2^{-x/2} \cdot \ln(2) - \lambda \cdot 2^{1-x} \ln(2)} \]

### Step 4: Substitute \(x = 2\) Again

Substituting \(x = 2\):

Numerator:

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

Denominator (with \(\lambda = -1\)):

\[ -\frac{1}{2} \cdot 2^{-2/2} \cdot \ln(2) - (-1) \cdot 2^{1-2} \cdot \ln(2) = -\frac{1}{2} \cdot \frac{1}{2} \ln(2) + \frac{1}{2} \ln(2) = -\frac{1}{4} \ln(2) + \frac{1}{2} \ln(2) = \frac{1}{4} \ln(2) \]

### Step 5: Final Calculation

Thus, the limit becomes:

\[ \frac{3 \ln(2)}{\frac{1}{4} \ln(2)} = 3 \cdot 4 = 12 \]

### Conclusion

The limit evaluates to \(12\) when \(\lambda = -1\).