`lim_(y to 0) ((y - 2) + 2sqrt(1 + y + y^(2)))/(2y)` is equal to _______

Answer

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

### Step 1: Substitute the limit
First, we substitute \( y = 0 \) directly into the expression:
\[ \frac{(0 - 2) + 2\sqrt{1 + 0 + 0^2}}{2 \cdot 0} = \frac{-2 + 2\sqrt{1}}{0} = \frac{-2 + 2}{0} = \frac{0}{0} \]
This gives us an indeterminate form \( \frac{0}{0} \), so we need to manipulate the expression.

### Step 2: Rationalize the numerator
To resolve the indeterminate form, we rationalize the numerator. We multiply the numerator and denominator by the conjugate of the expression involving the square root:
\[ \frac{(y - 2) + 2\sqrt{1 + y + y^2}}{2y} \cdot \frac{(y - 2) - 2\sqrt{1 + y + y^2}}{(y - 2) - 2\sqrt{1 + y + y^2}} \]
This gives us:
\[ \lim_{y \to 0} \frac{(y - 2)^2 - (2\sqrt{1 + y + y^2})^2}{2y \cdot ((y - 2) - 2\sqrt{1 + y + y^2})} \]

### Step 3: Simplify the numerator
Now, we simplify the numerator:
\[ (y - 2)^2 - 4(1 + y + y^2) = (y^2 - 4y + 4) - (4 + 4y + 4y^2) = y^2 - 4y + 4 - 4 - 4y - 4y^2 \]
This simplifies to:
\[ -y^2 - 8y \]

### Step 4: Substitute back into the limit
Now we substitute this back into our limit:
\[ \lim_{y \to 0} \frac{-y^2 - 8y}{2y \cdot ((y - 2) - 2\sqrt{1 + y + y^2})} \]
We can factor out \( -y \) from the numerator:
\[ \lim_{y \to 0} \frac{-y(y + 8)}{2y \cdot ((y - 2) - 2\sqrt{1 + y + y^2})} \]

### Step 5: Cancel \( y \)
Now we can cancel \( y \) from the numerator and denominator (as \( y \to 0 \)):
\[ \lim_{y \to 0} \frac{-(y + 8)}{2((y - 2) - 2\sqrt{1 + y + y^2})} \]

### Step 6: Evaluate the limit
Now we substitute \( y = 0 \):
\[ = \frac{-(0 + 8)}{2((-2) - 2\sqrt{1})} = \frac{-8}{2(-2 - 2)} = \frac{-8}{-8} = 1 \]

Thus, the limit is:
\[ \boxed{1} \]