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 \( x = -2 + h \)
We will substitute \( x \) with \( -2 + h \) where \( h \) approaches \( 0 \) from the positive side (since we are looking at the right-hand limit).

\[ \lim_{h \to 0^{+}} \frac{(-2 + h)^2 - 1}{2(-2 + h) + 4} \]

### Step 2: Simplify the numerator
Now we simplify the numerator:

\[ (-2 + h)^2 - 1 = (4 - 4h + h^2) - 1 = h^2 - 4h + 3 \]

### Step 3: Simplify the denominator
Now we simplify the denominator:

\[ 2(-2 + h) + 4 = -4 + 2h + 4 = 2h \]

### Step 4: Rewrite the limit
Now we can rewrite the limit as:

\[ \lim_{h \to 0^{+}} \frac{h^2 - 4h + 3}{2h} \]

### Step 5: Factor the numerator
Next, we factor the numerator \( h^2 - 4h + 3 \):

\[ h^2 - 4h + 3 = (h - 1)(h - 3) \]

### Step 6: Rewrite the limit again
Now we can rewrite the limit:

\[ \lim_{h \to 0^{+}} \frac{(h - 1)(h - 3)}{2h} \]

### Step 7: Evaluate the limit
Now we can evaluate the limit as \( h \) approaches \( 0 \):

\[ \frac{(-1)(-3)}{2 \cdot 0} = \frac{3}{0} \]

Since we are approaching from the right, \( 2h \) approaches \( 0^{+} \), thus \( \frac{3}{0^{+}} \) approaches \( +\infty \).

### Conclusion
Thus, the limit is:

\[ \lim_{x \to -2^{+}} \frac{x^2 - 1}{2x + 4} = +\infty \]