Evaluate `lim_(xto-2^(+)) (x^(2)-1)/(2x+4).`

Explanation

To evaluate the limit \( \lim_{x \to -2^+} \frac{x^2 - 1}{2x + 4} \), we will follow these steps:

### Step 1: Substitute \( x \) with \( -2 + h \)
We will rewrite the limit as \( h \) approaches 0, where \( h \) is a small positive number. Thus, we have:
\[ x = -2 + h \]
As \( x \to -2^+ \), \( h \to 0^+ \).

### Step 2: Rewrite the limit
Substituting \( x \) in the limit expression:
\[ \lim_{h \to 0^+} \frac{(-2 + h)^2 - 1}{2(-2 + h) + 4} \]

### Step 3: Simplify the numerator
Calculating the numerator:
\[ (-2 + h)^2 - 1 = (4 - 4h + h^2) - 1 = h^2 - 4h + 3 \]

### Step 4: Simplify the denominator
Calculating the denominator:
\[ 2(-2 + h) + 4 = -4 + 2h + 4 = 2h \]

### Step 5: Substitute back into the limit
Now substituting the simplified numerator and denominator back into the limit:
\[ \lim_{h \to 0^+} \frac{h^2 - 4h + 3}{2h} \]

### Step 6: Factor the numerator
We can factor the numerator:
\[ h^2 - 4h + 3 = (h - 1)(h - 3) \]
Thus, the limit becomes:
\[ \lim_{h \to 0^+} \frac{(h - 1)(h - 3)}{2h} \]

### Step 7: Evaluate the limit
As \( h \to 0^+ \):
- The numerator approaches \( (0 - 1)(0 - 3) = 3 \)
- The denominator approaches \( 2(0) = 0 \)

This implies:
\[ \lim_{h \to 0^+} \frac{3}{0} = \infty \]

### Conclusion
Thus, the limit is:
\[ \lim_{x \to -2^+} \frac{x^2 - 1}{2x + 4} = \infty \]

---