Step 1:

First, substituting the limit directly into the function, gives us:

[latex]\Large{\lim_{h\to -2}\frac{h^2+3h+2}{h+2} = \frac{(-2)^2 + 3(-2) + 2}{-2 + 2} = \frac{0}{0}}[/latex]

Hence, we have an indeterminate form, which means we must rewrite the function.

Step 2:

We notice that the numerator can be factored as follows:

[latex]\large{\lim_{h\to -2}\frac{h^2+3h+2}{h+2} = \lim_{h \to -2}\frac{(h+2)(h+1)}{h+2}}[/latex]

Now, we can cancel out the common terms:

[latex]\large{ = \lim_{h \to -2}\frac{h+1}{1} = \lim_{h \to -2}(h+1)}[/latex]

Step 3:

Now, we can plug in the limit once more, then solve:

[latex]\large{\lim_{h \to -2}h+1 = (-2) + 1 = -1}[/latex]

Solution Complete!