#### Answer

$-\dfrac{1}{2}$

#### Work Step by Step

We want to find $\lim\limits_{x \to -3}\dfrac{x+3}{x^2+4x+3}$, but we can't use the quotient rule for limits because the limit of the denominator as $x$ approaches -3 is zero, and if we try to substitute $x=-3$ directly into our function, we get zero in both the numerator and the denominator. This means we have a common factor of $x+3$ in both the numerator and the denominator.
Note that $\dfrac{x+3}{x^2+4x+3}=\dfrac{x+3}{(x+3)(x+1)}=\dfrac{1}{x+1}$ for all $x \neq -3.$
Thus $\lim\limits_{x \to -3}\dfrac{x+3}{x^2+4x+3}=\lim\limits_{x \to -3}\dfrac{1}{x+1}=\dfrac{1}{-3+1}=\dfrac{1}{-2}=-\dfrac{1}{2}.$